DualPivotQuicksort源码

x33g5p2x  于2021-12-30 转载在 其他  
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package java.util;

/** * This class implements the Dual-Pivot Quicksort algorithm by * Vladimir Yaroslavskiy, Jon Bentley, and Josh Bloch. The algorithm * offers O(n log(n)) performance on many data sets that cause other * quicksorts to degrade to quadratic performance, and is typically * faster than traditional (one-pivot) Quicksort implementations. * * All exposed methods are package-private, designed to be invoked * from public methods (in class Arrays) after performing any * necessary array bounds checks and expanding parameters into the * required forms. * * @author Vladimir Yaroslavskiy * @author Jon Bentley * @author Josh Bloch * * @version 2011.02.11 m765.827.12i:5\7pm * @since 1.7 */
final class DualPivotQuicksort {

    /** * Prevents instantiation. */
    private DualPivotQuicksort() {}

    /* * Tuning parameters. */

    /** * The maximum number of runs in merge sort. */
    private static final int MAX_RUN_COUNT = 67;

    /** * The maximum length of run in merge sort. */
    private static final int MAX_RUN_LENGTH = 33;

    /** * If the length of an array to be sorted is less than this * constant, Quicksort is used in preference to merge sort. */
    private static final int QUICKSORT_THRESHOLD = 286;

    /** * If the length of an array to be sorted is less than this * constant, insertion sort is used in preference to Quicksort. */
    private static final int INSERTION_SORT_THRESHOLD = 47;

    /** * If the length of a byte array to be sorted is greater than this * constant, counting sort is used in preference to insertion sort. */
    private static final int COUNTING_SORT_THRESHOLD_FOR_BYTE = 29;

    /** * If the length of a short or char array to be sorted is greater * than this constant, counting sort is used in preference to Quicksort. */
    private static final int COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR = 3200;

    /* * Sorting methods for seven primitive types. */

    /** * Sorts the specified range of the array using the given * workspace array slice if possible for merging * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    static void sort(int[] a, int left, int right,
                     int[] work, int workBase, int workLen) {
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
        }

        /* * Index run[i] is the start of i-th run * (ascending or descending sequence). */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) {
            if (a[k] < a[k + 1]) { // ascending
                while (++k <= right && a[k - 1] <= a[k]);
            } else if (a[k] > a[k + 1]) { // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                }
            } else { // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
                    if (--m == 0) {
                        sort(a, left, right, true);
                        return;
                    }
                }
            }

            /* * The array is not highly structured, * use Quicksort instead of merge sort. */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++) { // The last run contains one element
            run[++count] = right;
        } else if (count == 1) { // The array is already sorted
            return;
        }

        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        int[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) {
            work = new int[blen];
            workBase = 0;
        }
        if (odd == 0) {
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
        } else {
            b = work;
            ao = 0;
            bo = workBase - left;
        }

        // Merging
        for (int last; count > 1; count = last) {
            for (int k = (last = 0) + 2; k <= count; k += 2) {
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                        b[i + bo] = a[p++ + ao];
                    } else {
                        b[i + bo] = a[q++ + ao];
                    }
                }
                run[++last] = hi;
            }
            if ((count & 1) != 0) {
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            }
            int[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        }
    }

    /** * Sorts the specified range of the array by Dual-Pivot Quicksort. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param leftmost indicates if this part is the leftmost in the range */
    private static void sort(int[] a, int left, int right, boolean leftmost) {
        int length = right - left + 1;

        // Use insertion sort on tiny arrays
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {
                /* * Traditional (without sentinel) insertion sort, * optimized for server VM, is used in case of * the leftmost part. */
                for (int i = left, j = i; i < right; j = ++i) {
                    int ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }
            } else {
                /* * Skip the longest ascending sequence. */
                do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /* * Every element from adjoining part plays the role * of sentinel, therefore this allows us to avoid the * left range check on each iteration. Moreover, we use * the more optimized algorithm, so called pair insertion * sort, which is faster (in the context of Quicksort) * than traditional implementation of insertion sort. */
                for (int k = left; ++left <= right; k = ++left) {
                    int a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                int last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;
        }

        // Inexpensive approximation of length / 7
        int seventh = (length >> 3) + (length >> 6) + 1;

        /* * Sort five evenly spaced elements around (and including) the * center element in the range. These elements will be used for * pivot selection as described below. The choice for spacing * these elements was empirically determined to work well on * a wide variety of inputs. */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

        // Sort these elements using insertion sort
        if (a[e2] < a[e1]) { int t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { int t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { int t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { int t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

        // Pointers
        int less  = left;  // The index of the first element of center part
        int great = right; // The index before the first element of right part

        if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
            /* * Use the second and fourth of the five sorted elements as pivots. * These values are inexpensive approximations of the first and * second terciles of the array. Note that pivot1 <= pivot2. */
            int pivot1 = a[e2];
            int pivot2 = a[e4];

            /* * The first and the last elements to be sorted are moved to the * locations formerly occupied by the pivots. When partitioning * is complete, the pivots are swapped back into their final * positions, and excluded from subsequent sorting. */
            a[e2] = a[left];
            a[e4] = a[right];

            /* * Skip elements, which are less or greater than pivot values. */
            while (a[++less] < pivot1);
            while (a[--great] > pivot2);

            /* * Partitioning: * * left part center part right part * +--------------------------------------------------------------+ * | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 | * +--------------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot1 * pivot1 <= all in [less, k) <= pivot2 * all in (great, right) > pivot2 * * Pointer k is the first index of ?-part. */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                int ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /* * Here and below we use "a[i] = b; i++;" instead * of "a[i++] = b;" due to performance issue. */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /* * Here and below we use "a[i] = b; i--;" instead * of "a[i--] = b;" due to performance issue. */
                    a[great] = ak;
                    --great;
                }
            }

            // Swap pivots into their final positions
            a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
            a[right] = a[great + 1]; a[great + 1] = pivot2;

            // Sort left and right parts recursively, excluding known pivots
            sort(a, left, less - 2, leftmost);
            sort(a, great + 2, right, false);

            /* * If center part is too large (comprises > 4/7 of the array), * swap internal pivot values to ends. */
            if (less < e1 && e5 < great) {
                /* * Skip elements, which are equal to pivot values. */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /* * Partitioning: * * left part center part right part * +----------------------------------------------------------+ * | == pivot1 | pivot1 < && < pivot2 | ? | == pivot2 | * +----------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (*, less) == pivot1 * pivot1 < all in [less, k) < pivot2 * all in (great, *) == pivot2 * * Pointer k is the first index of ?-part. */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    int ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /* * Even though a[great] equals to pivot1, the * assignment a[less] = pivot1 may be incorrect, * if a[great] and pivot1 are floating-point zeros * of different signs. Therefore in float and * double sorting methods we have to use more * accurate assignment a[less] = a[great]. */
                            a[less] = pivot1;
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

        } else { // Partitioning with one pivot
            /* * Use the third of the five sorted elements as pivot. * This value is inexpensive approximation of the median. */
            int pivot = a[e3];

            /* * Partitioning degenerates to the traditional 3-way * (or "Dutch National Flag") schema: * * left part center part right part * +-------------------------------------------------+ * | < pivot | == pivot | ? | > pivot | * +-------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot * all in [less, k) == pivot * all in (great, right) > pivot * * Pointer k is the first index of ?-part. */
            for (int k = less; k <= great; ++k) {
                if (a[k] == pivot) {
                    continue;
                }
                int ak = a[k];
                if (ak < pivot) { // Move a[k] to left part
                    a[k] = a[less];
                    a[less] = ak;
                    ++less;
                } else { // a[k] > pivot - Move a[k] to right part
                    while (a[great] > pivot) {
                        --great;
                    }
                    if (a[great] < pivot) { // a[great] <= pivot
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // a[great] == pivot
                        /* * Even though a[great] equals to pivot, the * assignment a[k] = pivot may be incorrect, * if a[great] and pivot are floating-point * zeros of different signs. Therefore in float * and double sorting methods we have to use * more accurate assignment a[k] = a[great]. */
                        a[k] = pivot;
                    }
                    a[great] = ak;
                    --great;
                }
            }

            /* * Sort left and right parts recursively. * All elements from center part are equal * and, therefore, already sorted. */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }

    /** * Sorts the specified range of the array using the given * workspace array slice if possible for merging * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    static void sort(long[] a, int left, int right,
                     long[] work, int workBase, int workLen) {
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
        }

        /* * Index run[i] is the start of i-th run * (ascending or descending sequence). */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) {
            if (a[k] < a[k + 1]) { // ascending
                while (++k <= right && a[k - 1] <= a[k]);
            } else if (a[k] > a[k + 1]) { // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    long t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                }
            } else { // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
                    if (--m == 0) {
                        sort(a, left, right, true);
                        return;
                    }
                }
            }

            /* * The array is not highly structured, * use Quicksort instead of merge sort. */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++) { // The last run contains one element
            run[++count] = right;
        } else if (count == 1) { // The array is already sorted
            return;
        }

        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        long[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) {
            work = new long[blen];
            workBase = 0;
        }
        if (odd == 0) {
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
        } else {
            b = work;
            ao = 0;
            bo = workBase - left;
        }

        // Merging
        for (int last; count > 1; count = last) {
            for (int k = (last = 0) + 2; k <= count; k += 2) {
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                        b[i + bo] = a[p++ + ao];
                    } else {
                        b[i + bo] = a[q++ + ao];
                    }
                }
                run[++last] = hi;
            }
            if ((count & 1) != 0) {
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            }
            long[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        }
    }

    /** * Sorts the specified range of the array by Dual-Pivot Quicksort. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param leftmost indicates if this part is the leftmost in the range */
    private static void sort(long[] a, int left, int right, boolean leftmost) {
        int length = right - left + 1;

        // Use insertion sort on tiny arrays
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {
                /* * Traditional (without sentinel) insertion sort, * optimized for server VM, is used in case of * the leftmost part. */
                for (int i = left, j = i; i < right; j = ++i) {
                    long ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }
            } else {
                /* * Skip the longest ascending sequence. */
                do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /* * Every element from adjoining part plays the role * of sentinel, therefore this allows us to avoid the * left range check on each iteration. Moreover, we use * the more optimized algorithm, so called pair insertion * sort, which is faster (in the context of Quicksort) * than traditional implementation of insertion sort. */
                for (int k = left; ++left <= right; k = ++left) {
                    long a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                long last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;
        }

        // Inexpensive approximation of length / 7
        int seventh = (length >> 3) + (length >> 6) + 1;

        /* * Sort five evenly spaced elements around (and including) the * center element in the range. These elements will be used for * pivot selection as described below. The choice for spacing * these elements was empirically determined to work well on * a wide variety of inputs. */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

        // Sort these elements using insertion sort
        if (a[e2] < a[e1]) { long t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { long t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { long t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { long t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

        // Pointers
        int less  = left;  // The index of the first element of center part
        int great = right; // The index before the first element of right part

        if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
            /* * Use the second and fourth of the five sorted elements as pivots. * These values are inexpensive approximations of the first and * second terciles of the array. Note that pivot1 <= pivot2. */
            long pivot1 = a[e2];
            long pivot2 = a[e4];

            /* * The first and the last elements to be sorted are moved to the * locations formerly occupied by the pivots. When partitioning * is complete, the pivots are swapped back into their final * positions, and excluded from subsequent sorting. */
            a[e2] = a[left];
            a[e4] = a[right];

            /* * Skip elements, which are less or greater than pivot values. */
            while (a[++less] < pivot1);
            while (a[--great] > pivot2);

            /* * Partitioning: * * left part center part right part * +--------------------------------------------------------------+ * | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 | * +--------------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot1 * pivot1 <= all in [less, k) <= pivot2 * all in (great, right) > pivot2 * * Pointer k is the first index of ?-part. */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                long ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /* * Here and below we use "a[i] = b; i++;" instead * of "a[i++] = b;" due to performance issue. */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /* * Here and below we use "a[i] = b; i--;" instead * of "a[i--] = b;" due to performance issue. */
                    a[great] = ak;
                    --great;
                }
            }

            // Swap pivots into their final positions
            a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
            a[right] = a[great + 1]; a[great + 1] = pivot2;

            // Sort left and right parts recursively, excluding known pivots
            sort(a, left, less - 2, leftmost);
            sort(a, great + 2, right, false);

            /* * If center part is too large (comprises > 4/7 of the array), * swap internal pivot values to ends. */
            if (less < e1 && e5 < great) {
                /* * Skip elements, which are equal to pivot values. */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /* * Partitioning: * * left part center part right part * +----------------------------------------------------------+ * | == pivot1 | pivot1 < && < pivot2 | ? | == pivot2 | * +----------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (*, less) == pivot1 * pivot1 < all in [less, k) < pivot2 * all in (great, *) == pivot2 * * Pointer k is the first index of ?-part. */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    long ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /* * Even though a[great] equals to pivot1, the * assignment a[less] = pivot1 may be incorrect, * if a[great] and pivot1 are floating-point zeros * of different signs. Therefore in float and * double sorting methods we have to use more * accurate assignment a[less] = a[great]. */
                            a[less] = pivot1;
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

        } else { // Partitioning with one pivot
            /* * Use the third of the five sorted elements as pivot. * This value is inexpensive approximation of the median. */
            long pivot = a[e3];

            /* * Partitioning degenerates to the traditional 3-way * (or "Dutch National Flag") schema: * * left part center part right part * +-------------------------------------------------+ * | < pivot | == pivot | ? | > pivot | * +-------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot * all in [less, k) == pivot * all in (great, right) > pivot * * Pointer k is the first index of ?-part. */
            for (int k = less; k <= great; ++k) {
                if (a[k] == pivot) {
                    continue;
                }
                long ak = a[k];
                if (ak < pivot) { // Move a[k] to left part
                    a[k] = a[less];
                    a[less] = ak;
                    ++less;
                } else { // a[k] > pivot - Move a[k] to right part
                    while (a[great] > pivot) {
                        --great;
                    }
                    if (a[great] < pivot) { // a[great] <= pivot
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // a[great] == pivot
                        /* * Even though a[great] equals to pivot, the * assignment a[k] = pivot may be incorrect, * if a[great] and pivot are floating-point * zeros of different signs. Therefore in float * and double sorting methods we have to use * more accurate assignment a[k] = a[great]. */
                        a[k] = pivot;
                    }
                    a[great] = ak;
                    --great;
                }
            }

            /* * Sort left and right parts recursively. * All elements from center part are equal * and, therefore, already sorted. */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }

    /** * Sorts the specified range of the array using the given * workspace array slice if possible for merging * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    static void sort(short[] a, int left, int right,
                     short[] work, int workBase, int workLen) {
        // Use counting sort on large arrays
        if (right - left > COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR) {
            int[] count = new int[NUM_SHORT_VALUES];

            for (int i = left - 1; ++i <= right;
                count[a[i] - Short.MIN_VALUE]++
            );
            for (int i = NUM_SHORT_VALUES, k = right + 1; k > left; ) {
                while (count[--i] == 0);
                short value = (short) (i + Short.MIN_VALUE);
                int s = count[i];

                do {
                    a[--k] = value;
                } while (--s > 0);
            }
        } else { // Use Dual-Pivot Quicksort on small arrays
            doSort(a, left, right, work, workBase, workLen);
        }
    }

    /** The number of distinct short values. */
    private static final int NUM_SHORT_VALUES = 1 << 16;

    /** * Sorts the specified range of the array. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    private static void doSort(short[] a, int left, int right,
                               short[] work, int workBase, int workLen) {
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
        }

        /* * Index run[i] is the start of i-th run * (ascending or descending sequence). */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) {
            if (a[k] < a[k + 1]) { // ascending
                while (++k <= right && a[k - 1] <= a[k]);
            } else if (a[k] > a[k + 1]) { // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    short t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                }
            } else { // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
                    if (--m == 0) {
                        sort(a, left, right, true);
                        return;
                    }
                }
            }

            /* * The array is not highly structured, * use Quicksort instead of merge sort. */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++) { // The last run contains one element
            run[++count] = right;
        } else if (count == 1) { // The array is already sorted
            return;
        }

        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        short[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) {
            work = new short[blen];
            workBase = 0;
        }
        if (odd == 0) {
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
        } else {
            b = work;
            ao = 0;
            bo = workBase - left;
        }

        // Merging
        for (int last; count > 1; count = last) {
            for (int k = (last = 0) + 2; k <= count; k += 2) {
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                        b[i + bo] = a[p++ + ao];
                    } else {
                        b[i + bo] = a[q++ + ao];
                    }
                }
                run[++last] = hi;
            }
            if ((count & 1) != 0) {
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            }
            short[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        }
    }

    /** * Sorts the specified range of the array by Dual-Pivot Quicksort. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param leftmost indicates if this part is the leftmost in the range */
    private static void sort(short[] a, int left, int right, boolean leftmost) {
        int length = right - left + 1;

        // Use insertion sort on tiny arrays
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {
                /* * Traditional (without sentinel) insertion sort, * optimized for server VM, is used in case of * the leftmost part. */
                for (int i = left, j = i; i < right; j = ++i) {
                    short ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }
            } else {
                /* * Skip the longest ascending sequence. */
                do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /* * Every element from adjoining part plays the role * of sentinel, therefore this allows us to avoid the * left range check on each iteration. Moreover, we use * the more optimized algorithm, so called pair insertion * sort, which is faster (in the context of Quicksort) * than traditional implementation of insertion sort. */
                for (int k = left; ++left <= right; k = ++left) {
                    short a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                short last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;
        }

        // Inexpensive approximation of length / 7
        int seventh = (length >> 3) + (length >> 6) + 1;

        /* * Sort five evenly spaced elements around (and including) the * center element in the range. These elements will be used for * pivot selection as described below. The choice for spacing * these elements was empirically determined to work well on * a wide variety of inputs. */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

        // Sort these elements using insertion sort
        if (a[e2] < a[e1]) { short t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { short t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { short t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { short t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

        // Pointers
        int less  = left;  // The index of the first element of center part
        int great = right; // The index before the first element of right part

        if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
            /* * Use the second and fourth of the five sorted elements as pivots. * These values are inexpensive approximations of the first and * second terciles of the array. Note that pivot1 <= pivot2. */
            short pivot1 = a[e2];
            short pivot2 = a[e4];

            /* * The first and the last elements to be sorted are moved to the * locations formerly occupied by the pivots. When partitioning * is complete, the pivots are swapped back into their final * positions, and excluded from subsequent sorting. */
            a[e2] = a[left];
            a[e4] = a[right];

            /* * Skip elements, which are less or greater than pivot values. */
            while (a[++less] < pivot1);
            while (a[--great] > pivot2);

            /* * Partitioning: * * left part center part right part * +--------------------------------------------------------------+ * | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 | * +--------------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot1 * pivot1 <= all in [less, k) <= pivot2 * all in (great, right) > pivot2 * * Pointer k is the first index of ?-part. */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                short ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /* * Here and below we use "a[i] = b; i++;" instead * of "a[i++] = b;" due to performance issue. */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /* * Here and below we use "a[i] = b; i--;" instead * of "a[i--] = b;" due to performance issue. */
                    a[great] = ak;
                    --great;
                }
            }

            // Swap pivots into their final positions
            a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
            a[right] = a[great + 1]; a[great + 1] = pivot2;

            // Sort left and right parts recursively, excluding known pivots
            sort(a, left, less - 2, leftmost);
            sort(a, great + 2, right, false);

            /* * If center part is too large (comprises > 4/7 of the array), * swap internal pivot values to ends. */
            if (less < e1 && e5 < great) {
                /* * Skip elements, which are equal to pivot values. */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /* * Partitioning: * * left part center part right part * +----------------------------------------------------------+ * | == pivot1 | pivot1 < && < pivot2 | ? | == pivot2 | * +----------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (*, less) == pivot1 * pivot1 < all in [less, k) < pivot2 * all in (great, *) == pivot2 * * Pointer k is the first index of ?-part. */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    short ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /* * Even though a[great] equals to pivot1, the * assignment a[less] = pivot1 may be incorrect, * if a[great] and pivot1 are floating-point zeros * of different signs. Therefore in float and * double sorting methods we have to use more * accurate assignment a[less] = a[great]. */
                            a[less] = pivot1;
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

        } else { // Partitioning with one pivot
            /* * Use the third of the five sorted elements as pivot. * This value is inexpensive approximation of the median. */
            short pivot = a[e3];

            /* * Partitioning degenerates to the traditional 3-way * (or "Dutch National Flag") schema: * * left part center part right part * +-------------------------------------------------+ * | < pivot | == pivot | ? | > pivot | * +-------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot * all in [less, k) == pivot * all in (great, right) > pivot * * Pointer k is the first index of ?-part. */
            for (int k = less; k <= great; ++k) {
                if (a[k] == pivot) {
                    continue;
                }
                short ak = a[k];
                if (ak < pivot) { // Move a[k] to left part
                    a[k] = a[less];
                    a[less] = ak;
                    ++less;
                } else { // a[k] > pivot - Move a[k] to right part
                    while (a[great] > pivot) {
                        --great;
                    }
                    if (a[great] < pivot) { // a[great] <= pivot
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // a[great] == pivot
                        /* * Even though a[great] equals to pivot, the * assignment a[k] = pivot may be incorrect, * if a[great] and pivot are floating-point * zeros of different signs. Therefore in float * and double sorting methods we have to use * more accurate assignment a[k] = a[great]. */
                        a[k] = pivot;
                    }
                    a[great] = ak;
                    --great;
                }
            }

            /* * Sort left and right parts recursively. * All elements from center part are equal * and, therefore, already sorted. */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }

    /** * Sorts the specified range of the array using the given * workspace array slice if possible for merging * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    static void sort(char[] a, int left, int right,
                     char[] work, int workBase, int workLen) {
        // Use counting sort on large arrays
        if (right - left > COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR) {
            int[] count = new int[NUM_CHAR_VALUES];

            for (int i = left - 1; ++i <= right;
                count[a[i]]++
            );
            for (int i = NUM_CHAR_VALUES, k = right + 1; k > left; ) {
                while (count[--i] == 0);
                char value = (char) i;
                int s = count[i];

                do {
                    a[--k] = value;
                } while (--s > 0);
            }
        } else { // Use Dual-Pivot Quicksort on small arrays
            doSort(a, left, right, work, workBase, workLen);
        }
    }

    /** The number of distinct char values. */
    private static final int NUM_CHAR_VALUES = 1 << 16;

    /** * Sorts the specified range of the array. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    private static void doSort(char[] a, int left, int right,
                               char[] work, int workBase, int workLen) {
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
        }

        /* * Index run[i] is the start of i-th run * (ascending or descending sequence). */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) {
            if (a[k] < a[k + 1]) { // ascending
                while (++k <= right && a[k - 1] <= a[k]);
            } else if (a[k] > a[k + 1]) { // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    char t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                }
            } else { // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
                    if (--m == 0) {
                        sort(a, left, right, true);
                        return;
                    }
                }
            }

            /* * The array is not highly structured, * use Quicksort instead of merge sort. */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++) { // The last run contains one element
            run[++count] = right;
        } else if (count == 1) { // The array is already sorted
            return;
        }

        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        char[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) {
            work = new char[blen];
            workBase = 0;
        }
        if (odd == 0) {
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
        } else {
            b = work;
            ao = 0;
            bo = workBase - left;
        }

        // Merging
        for (int last; count > 1; count = last) {
            for (int k = (last = 0) + 2; k <= count; k += 2) {
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                        b[i + bo] = a[p++ + ao];
                    } else {
                        b[i + bo] = a[q++ + ao];
                    }
                }
                run[++last] = hi;
            }
            if ((count & 1) != 0) {
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            }
            char[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        }
    }

    /** * Sorts the specified range of the array by Dual-Pivot Quicksort. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param leftmost indicates if this part is the leftmost in the range */
    private static void sort(char[] a, int left, int right, boolean leftmost) {
        int length = right - left + 1;

        // Use insertion sort on tiny arrays
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {
                /* * Traditional (without sentinel) insertion sort, * optimized for server VM, is used in case of * the leftmost part. */
                for (int i = left, j = i; i < right; j = ++i) {
                    char ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }
            } else {
                /* * Skip the longest ascending sequence. */
                do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /* * Every element from adjoining part plays the role * of sentinel, therefore this allows us to avoid the * left range check on each iteration. Moreover, we use * the more optimized algorithm, so called pair insertion * sort, which is faster (in the context of Quicksort) * than traditional implementation of insertion sort. */
                for (int k = left; ++left <= right; k = ++left) {
                    char a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                char last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;
        }

        // Inexpensive approximation of length / 7
        int seventh = (length >> 3) + (length >> 6) + 1;

        /* * Sort five evenly spaced elements around (and including) the * center element in the range. These elements will be used for * pivot selection as described below. The choice for spacing * these elements was empirically determined to work well on * a wide variety of inputs. */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

        // Sort these elements using insertion sort
        if (a[e2] < a[e1]) { char t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { char t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { char t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { char t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

        // Pointers
        int less  = left;  // The index of the first element of center part
        int great = right; // The index before the first element of right part

        if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
            /* * Use the second and fourth of the five sorted elements as pivots. * These values are inexpensive approximations of the first and * second terciles of the array. Note that pivot1 <= pivot2. */
            char pivot1 = a[e2];
            char pivot2 = a[e4];

            /* * The first and the last elements to be sorted are moved to the * locations formerly occupied by the pivots. When partitioning * is complete, the pivots are swapped back into their final * positions, and excluded from subsequent sorting. */
            a[e2] = a[left];
            a[e4] = a[right];

            /* * Skip elements, which are less or greater than pivot values. */
            while (a[++less] < pivot1);
            while (a[--great] > pivot2);

            /* * Partitioning: * * left part center part right part * +--------------------------------------------------------------+ * | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 | * +--------------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot1 * pivot1 <= all in [less, k) <= pivot2 * all in (great, right) > pivot2 * * Pointer k is the first index of ?-part. */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                char ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /* * Here and below we use "a[i] = b; i++;" instead * of "a[i++] = b;" due to performance issue. */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /* * Here and below we use "a[i] = b; i--;" instead * of "a[i--] = b;" due to performance issue. */
                    a[great] = ak;
                    --great;
                }
            }

            // Swap pivots into their final positions
            a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
            a[right] = a[great + 1]; a[great + 1] = pivot2;

            // Sort left and right parts recursively, excluding known pivots
            sort(a, left, less - 2, leftmost);
            sort(a, great + 2, right, false);

            /* * If center part is too large (comprises > 4/7 of the array), * swap internal pivot values to ends. */
            if (less < e1 && e5 < great) {
                /* * Skip elements, which are equal to pivot values. */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /* * Partitioning: * * left part center part right part * +----------------------------------------------------------+ * | == pivot1 | pivot1 < && < pivot2 | ? | == pivot2 | * +----------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (*, less) == pivot1 * pivot1 < all in [less, k) < pivot2 * all in (great, *) == pivot2 * * Pointer k is the first index of ?-part. */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    char ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /* * Even though a[great] equals to pivot1, the * assignment a[less] = pivot1 may be incorrect, * if a[great] and pivot1 are floating-point zeros * of different signs. Therefore in float and * double sorting methods we have to use more * accurate assignment a[less] = a[great]. */
                            a[less] = pivot1;
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

        } else { // Partitioning with one pivot
            /* * Use the third of the five sorted elements as pivot. * This value is inexpensive approximation of the median. */
            char pivot = a[e3];

            /* * Partitioning degenerates to the traditional 3-way * (or "Dutch National Flag") schema: * * left part center part right part * +-------------------------------------------------+ * | < pivot | == pivot | ? | > pivot | * +-------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot * all in [less, k) == pivot * all in (great, right) > pivot * * Pointer k is the first index of ?-part. */
            for (int k = less; k <= great; ++k) {
                if (a[k] == pivot) {
                    continue;
                }
                char ak = a[k];
                if (ak < pivot) { // Move a[k] to left part
                    a[k] = a[less];
                    a[less] = ak;
                    ++less;
                } else { // a[k] > pivot - Move a[k] to right part
                    while (a[great] > pivot) {
                        --great;
                    }
                    if (a[great] < pivot) { // a[great] <= pivot
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // a[great] == pivot
                        /* * Even though a[great] equals to pivot, the * assignment a[k] = pivot may be incorrect, * if a[great] and pivot are floating-point * zeros of different signs. Therefore in float * and double sorting methods we have to use * more accurate assignment a[k] = a[great]. */
                        a[k] = pivot;
                    }
                    a[great] = ak;
                    --great;
                }
            }

            /* * Sort left and right parts recursively. * All elements from center part are equal * and, therefore, already sorted. */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }

    /** The number of distinct byte values. */
    private static final int NUM_BYTE_VALUES = 1 << 8;

    /** * Sorts the specified range of the array. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted */
    static void sort(byte[] a, int left, int right) {
        // Use counting sort on large arrays
        if (right - left > COUNTING_SORT_THRESHOLD_FOR_BYTE) {
            int[] count = new int[NUM_BYTE_VALUES];

            for (int i = left - 1; ++i <= right;
                count[a[i] - Byte.MIN_VALUE]++
            );
            for (int i = NUM_BYTE_VALUES, k = right + 1; k > left; ) {
                while (count[--i] == 0);
                byte value = (byte) (i + Byte.MIN_VALUE);
                int s = count[i];

                do {
                    a[--k] = value;
                } while (--s > 0);
            }
        } else { // Use insertion sort on small arrays
            for (int i = left, j = i; i < right; j = ++i) {
                byte ai = a[i + 1];
                while (ai < a[j]) {
                    a[j + 1] = a[j];
                    if (j-- == left) {
                        break;
                    }
                }
                a[j + 1] = ai;
            }
        }
    }

    /** * Sorts the specified range of the array using the given * workspace array slice if possible for merging * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    static void sort(float[] a, int left, int right,
                     float[] work, int workBase, int workLen) {
        /* * Phase 1: Move NaNs to the end of the array. */
        while (left <= right && Float.isNaN(a[right])) {
            --right;
        }
        for (int k = right; --k >= left; ) {
            float ak = a[k];
            if (ak != ak) { // a[k] is NaN
                a[k] = a[right];
                a[right] = ak;
                --right;
            }
        }

        /* * Phase 2: Sort everything except NaNs (which are already in place). */
        doSort(a, left, right, work, workBase, workLen);

        /* * Phase 3: Place negative zeros before positive zeros. */
        int hi = right;

        /* * Find the first zero, or first positive, or last negative element. */
        while (left < hi) {
            int middle = (left + hi) >>> 1;
            float middleValue = a[middle];

            if (middleValue < 0.0f) {
                left = middle + 1;
            } else {
                hi = middle;
            }
        }

        /* * Skip the last negative value (if any) or all leading negative zeros. */
        while (left <= right && Float.floatToRawIntBits(a[left]) < 0) {
            ++left;
        }

        /* * Move negative zeros to the beginning of the sub-range. * * Partitioning: * * +----------------------------------------------------+ * | < 0.0 | -0.0 | 0.0 | ? ( >= 0.0 ) | * +----------------------------------------------------+ * ^ ^ ^ * | | | * left p k * * Invariants: * * all in (*, left) < 0.0 * all in [left, p) == -0.0 * all in [p, k) == 0.0 * all in [k, right] >= 0.0 * * Pointer k is the first index of ?-part. */
        for (int k = left, p = left - 1; ++k <= right; ) {
            float ak = a[k];
            if (ak != 0.0f) {
                break;
            }
            if (Float.floatToRawIntBits(ak) < 0) { // ak is -0.0f
                a[k] = 0.0f;
                a[++p] = -0.0f;
            }
        }
    }

    /** * Sorts the specified range of the array. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    private static void doSort(float[] a, int left, int right,
                               float[] work, int workBase, int workLen) {
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
        }

        /* * Index run[i] is the start of i-th run * (ascending or descending sequence). */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) {
            if (a[k] < a[k + 1]) { // ascending
                while (++k <= right && a[k - 1] <= a[k]);
            } else if (a[k] > a[k + 1]) { // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    float t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                }
            } else { // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
                    if (--m == 0) {
                        sort(a, left, right, true);
                        return;
                    }
                }
            }

            /* * The array is not highly structured, * use Quicksort instead of merge sort. */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++) { // The last run contains one element
            run[++count] = right;
        } else if (count == 1) { // The array is already sorted
            return;
        }

        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        float[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) {
            work = new float[blen];
            workBase = 0;
        }
        if (odd == 0) {
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
        } else {
            b = work;
            ao = 0;
            bo = workBase - left;
        }

        // Merging
        for (int last; count > 1; count = last) {
            for (int k = (last = 0) + 2; k <= count; k += 2) {
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                        b[i + bo] = a[p++ + ao];
                    } else {
                        b[i + bo] = a[q++ + ao];
                    }
                }
                run[++last] = hi;
            }
            if ((count & 1) != 0) {
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            }
            float[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        }
    }

    /** * Sorts the specified range of the array by Dual-Pivot Quicksort. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param leftmost indicates if this part is the leftmost in the range */
    private static void sort(float[] a, int left, int right, boolean leftmost) {
        int length = right - left + 1;

        // Use insertion sort on tiny arrays
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {
                /* * Traditional (without sentinel) insertion sort, * optimized for server VM, is used in case of * the leftmost part. */
                for (int i = left, j = i; i < right; j = ++i) {
                    float ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }
            } else {
                /* * Skip the longest ascending sequence. */
                do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /* * Every element from adjoining part plays the role * of sentinel, therefore this allows us to avoid the * left range check on each iteration. Moreover, we use * the more optimized algorithm, so called pair insertion * sort, which is faster (in the context of Quicksort) * than traditional implementation of insertion sort. */
                for (int k = left; ++left <= right; k = ++left) {
                    float a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                float last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;
        }

        // Inexpensive approximation of length / 7
        int seventh = (length >> 3) + (length >> 6) + 1;

        /* * Sort five evenly spaced elements around (and including) the * center element in the range. These elements will be used for * pivot selection as described below. The choice for spacing * these elements was empirically determined to work well on * a wide variety of inputs. */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

        // Sort these elements using insertion sort
        if (a[e2] < a[e1]) { float t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { float t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { float t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { float t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

        // Pointers
        int less  = left;  // The index of the first element of center part
        int great = right; // The index before the first element of right part

        if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
            /* * Use the second and fourth of the five sorted elements as pivots. * These values are inexpensive approximations of the first and * second terciles of the array. Note that pivot1 <= pivot2. */
            float pivot1 = a[e2];
            float pivot2 = a[e4];

            /* * The first and the last elements to be sorted are moved to the * locations formerly occupied by the pivots. When partitioning * is complete, the pivots are swapped back into their final * positions, and excluded from subsequent sorting. */
            a[e2] = a[left];
            a[e4] = a[right];

            /* * Skip elements, which are less or greater than pivot values. */
            while (a[++less] < pivot1);
            while (a[--great] > pivot2);

            /* * Partitioning: * * left part center part right part * +--------------------------------------------------------------+ * | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 | * +--------------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot1 * pivot1 <= all in [less, k) <= pivot2 * all in (great, right) > pivot2 * * Pointer k is the first index of ?-part. */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                float ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /* * Here and below we use "a[i] = b; i++;" instead * of "a[i++] = b;" due to performance issue. */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /* * Here and below we use "a[i] = b; i--;" instead * of "a[i--] = b;" due to performance issue. */
                    a[great] = ak;
                    --great;
                }
            }

            // Swap pivots into their final positions
            a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
            a[right] = a[great + 1]; a[great + 1] = pivot2;

            // Sort left and right parts recursively, excluding known pivots
            sort(a, left, less - 2, leftmost);
            sort(a, great + 2, right, false);

            /* * If center part is too large (comprises > 4/7 of the array), * swap internal pivot values to ends. */
            if (less < e1 && e5 < great) {
                /* * Skip elements, which are equal to pivot values. */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /* * Partitioning: * * left part center part right part * +----------------------------------------------------------+ * | == pivot1 | pivot1 < && < pivot2 | ? | == pivot2 | * +----------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (*, less) == pivot1 * pivot1 < all in [less, k) < pivot2 * all in (great, *) == pivot2 * * Pointer k is the first index of ?-part. */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    float ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /* * Even though a[great] equals to pivot1, the * assignment a[less] = pivot1 may be incorrect, * if a[great] and pivot1 are floating-point zeros * of different signs. Therefore in float and * double sorting methods we have to use more * accurate assignment a[less] = a[great]. */
                            a[less] = a[great];
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

        } else { // Partitioning with one pivot
            /* * Use the third of the five sorted elements as pivot. * This value is inexpensive approximation of the median. */
            float pivot = a[e3];

            /* * Partitioning degenerates to the traditional 3-way * (or "Dutch National Flag") schema: * * left part center part right part * +-------------------------------------------------+ * | < pivot | == pivot | ? | > pivot | * +-------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot * all in [less, k) == pivot * all in (great, right) > pivot * * Pointer k is the first index of ?-part. */
            for (int k = less; k <= great; ++k) {
                if (a[k] == pivot) {
                    continue;
                }
                float ak = a[k];
                if (ak < pivot) { // Move a[k] to left part
                    a[k] = a[less];
                    a[less] = ak;
                    ++less;
                } else { // a[k] > pivot - Move a[k] to right part
                    while (a[great] > pivot) {
                        --great;
                    }
                    if (a[great] < pivot) { // a[great] <= pivot
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // a[great] == pivot
                        /* * Even though a[great] equals to pivot, the * assignment a[k] = pivot may be incorrect, * if a[great] and pivot are floating-point * zeros of different signs. Therefore in float * and double sorting methods we have to use * more accurate assignment a[k] = a[great]. */
                        a[k] = a[great];
                    }
                    a[great] = ak;
                    --great;
                }
            }

            /* * Sort left and right parts recursively. * All elements from center part are equal * and, therefore, already sorted. */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }

    /** * Sorts the specified range of the array using the given * workspace array slice if possible for merging * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    static void sort(double[] a, int left, int right,
                     double[] work, int workBase, int workLen) {
        /* * Phase 1: Move NaNs to the end of the array. */
        while (left <= right && Double.isNaN(a[right])) {
            --right;
        }
        for (int k = right; --k >= left; ) {
            double ak = a[k];
            if (ak != ak) { // a[k] is NaN
                a[k] = a[right];
                a[right] = ak;
                --right;
            }
        }

        /* * Phase 2: Sort everything except NaNs (which are already in place). */
        doSort(a, left, right, work, workBase, workLen);

        /* * Phase 3: Place negative zeros before positive zeros. */
        int hi = right;

        /* * Find the first zero, or first positive, or last negative element. */
        while (left < hi) {
            int middle = (left + hi) >>> 1;
            double middleValue = a[middle];

            if (middleValue < 0.0d) {
                left = middle + 1;
            } else {
                hi = middle;
            }
        }

        /* * Skip the last negative value (if any) or all leading negative zeros. */
        while (left <= right && Double.doubleToRawLongBits(a[left]) < 0) {
            ++left;
        }

        /* * Move negative zeros to the beginning of the sub-range. * * Partitioning: * * +----------------------------------------------------+ * | < 0.0 | -0.0 | 0.0 | ? ( >= 0.0 ) | * +----------------------------------------------------+ * ^ ^ ^ * | | | * left p k * * Invariants: * * all in (*, left) < 0.0 * all in [left, p) == -0.0 * all in [p, k) == 0.0 * all in [k, right] >= 0.0 * * Pointer k is the first index of ?-part. */
        for (int k = left, p = left - 1; ++k <= right; ) {
            double ak = a[k];
            if (ak != 0.0d) {
                break;
            }
            if (Double.doubleToRawLongBits(ak) < 0) { // ak is -0.0d
                a[k] = 0.0d;
                a[++p] = -0.0d;
            }
        }
    }

    /** * Sorts the specified range of the array. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array */
    private static void doSort(double[] a, int left, int right,
                               double[] work, int workBase, int workLen) {
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
        }

        /* * Index run[i] is the start of i-th run * (ascending or descending sequence). */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) {
            if (a[k] < a[k + 1]) { // ascending
                while (++k <= right && a[k - 1] <= a[k]);
            } else if (a[k] > a[k + 1]) { // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    double t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                }
            } else { // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
                    if (--m == 0) {
                        sort(a, left, right, true);
                        return;
                    }
                }
            }

            /* * The array is not highly structured, * use Quicksort instead of merge sort. */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++) { // The last run contains one element
            run[++count] = right;
        } else if (count == 1) { // The array is already sorted
            return;
        }

        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        double[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) {
            work = new double[blen];
            workBase = 0;
        }
        if (odd == 0) {
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
        } else {
            b = work;
            ao = 0;
            bo = workBase - left;
        }

        // Merging
        for (int last; count > 1; count = last) {
            for (int k = (last = 0) + 2; k <= count; k += 2) {
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                        b[i + bo] = a[p++ + ao];
                    } else {
                        b[i + bo] = a[q++ + ao];
                    }
                }
                run[++last] = hi;
            }
            if ((count & 1) != 0) {
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            }
            double[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        }
    }

    /** * Sorts the specified range of the array by Dual-Pivot Quicksort. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param leftmost indicates if this part is the leftmost in the range */
    private static void sort(double[] a, int left, int right, boolean leftmost) {
        int length = right - left + 1;

        // Use insertion sort on tiny arrays
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {
                /* * Traditional (without sentinel) insertion sort, * optimized for server VM, is used in case of * the leftmost part. */
                for (int i = left, j = i; i < right; j = ++i) {
                    double ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }
            } else {
                /* * Skip the longest ascending sequence. */
                do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /* * Every element from adjoining part plays the role * of sentinel, therefore this allows us to avoid the * left range check on each iteration. Moreover, we use * the more optimized algorithm, so called pair insertion * sort, which is faster (in the context of Quicksort) * than traditional implementation of insertion sort. */
                for (int k = left; ++left <= right; k = ++left) {
                    double a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                double last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;
        }

        // Inexpensive approximation of length / 7
        int seventh = (length >> 3) + (length >> 6) + 1;

        /* * Sort five evenly spaced elements around (and including) the * center element in the range. These elements will be used for * pivot selection as described below. The choice for spacing * these elements was empirically determined to work well on * a wide variety of inputs. */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

        // Sort these elements using insertion sort
        if (a[e2] < a[e1]) { double t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { double t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { double t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { double t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

        // Pointers
        int less  = left;  // The index of the first element of center part
        int great = right; // The index before the first element of right part

        if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
            /* * Use the second and fourth of the five sorted elements as pivots. * These values are inexpensive approximations of the first and * second terciles of the array. Note that pivot1 <= pivot2. */
            double pivot1 = a[e2];
            double pivot2 = a[e4];

            /* * The first and the last elements to be sorted are moved to the * locations formerly occupied by the pivots. When partitioning * is complete, the pivots are swapped back into their final * positions, and excluded from subsequent sorting. */
            a[e2] = a[left];
            a[e4] = a[right];

            /* * Skip elements, which are less or greater than pivot values. */
            while (a[++less] < pivot1);
            while (a[--great] > pivot2);

            /* * Partitioning: * * left part center part right part * +--------------------------------------------------------------+ * | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 | * +--------------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot1 * pivot1 <= all in [less, k) <= pivot2 * all in (great, right) > pivot2 * * Pointer k is the first index of ?-part. */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                double ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /* * Here and below we use "a[i] = b; i++;" instead * of "a[i++] = b;" due to performance issue. */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /* * Here and below we use "a[i] = b; i--;" instead * of "a[i--] = b;" due to performance issue. */
                    a[great] = ak;
                    --great;
                }
            }

            // Swap pivots into their final positions
            a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
            a[right] = a[great + 1]; a[great + 1] = pivot2;

            // Sort left and right parts recursively, excluding known pivots
            sort(a, left, less - 2, leftmost);
            sort(a, great + 2, right, false);

            /* * If center part is too large (comprises > 4/7 of the array), * swap internal pivot values to ends. */
            if (less < e1 && e5 < great) {
                /* * Skip elements, which are equal to pivot values. */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /* * Partitioning: * * left part center part right part * +----------------------------------------------------------+ * | == pivot1 | pivot1 < && < pivot2 | ? | == pivot2 | * +----------------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (*, less) == pivot1 * pivot1 < all in [less, k) < pivot2 * all in (great, *) == pivot2 * * Pointer k is the first index of ?-part. */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    double ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /* * Even though a[great] equals to pivot1, the * assignment a[less] = pivot1 may be incorrect, * if a[great] and pivot1 are floating-point zeros * of different signs. Therefore in float and * double sorting methods we have to use more * accurate assignment a[less] = a[great]. */
                            a[less] = a[great];
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

        } else { // Partitioning with one pivot
            /* * Use the third of the five sorted elements as pivot. * This value is inexpensive approximation of the median. */
            double pivot = a[e3];

            /* * Partitioning degenerates to the traditional 3-way * (or "Dutch National Flag") schema: * * left part center part right part * +-------------------------------------------------+ * | < pivot | == pivot | ? | > pivot | * +-------------------------------------------------+ * ^ ^ ^ * | | | * less k great * * Invariants: * * all in (left, less) < pivot * all in [less, k) == pivot * all in (great, right) > pivot * * Pointer k is the first index of ?-part. */
            for (int k = less; k <= great; ++k) {
                if (a[k] == pivot) {
                    continue;
                }
                double ak = a[k];
                if (ak < pivot) { // Move a[k] to left part
                    a[k] = a[less];
                    a[less] = ak;
                    ++less;
                } else { // a[k] > pivot - Move a[k] to right part
                    while (a[great] > pivot) {
                        --great;
                    }
                    if (a[great] < pivot) { // a[great] <= pivot
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // a[great] == pivot
                        /* * Even though a[great] equals to pivot, the * assignment a[k] = pivot may be incorrect, * if a[great] and pivot are floating-point * zeros of different signs. Therefore in float * and double sorting methods we have to use * more accurate assignment a[k] = a[great]. */
                        a[k] = a[great];
                    }
                    a[great] = ak;
                    --great;
                }
            }

            /* * Sort left and right parts recursively. * All elements from center part are equal * and, therefore, already sorted. */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }
}

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