归并排序及其应用场景

x33g5p2x  于11个月前 转载在 其他  
字(1.5k)|赞(0)|评价(0)|浏览(97)

一、归并排序的概念

二、归并排序递归与非递归实现

归并排序递归实现,分治为每个区间元素都有序,那么就得把区间 分治成为1才能保证区间中每个元素都有序,在归并:

//归并排序
void _MergeSort(int*arr, int left, int right, int* tmp)
{
	if (left >= right)
	{
		return;
	}
	int mid = left + (right - left) / 2;
	_MergeSort(arr, left, mid, tmp);
	_MergeSort(arr, mid + 1, right, tmp);

	int begin1 = left;
	int end1 = mid;

	int begin2 = mid + 1;
	int end2 = right;

	int i = left; 

	while (begin1 <= end1&&begin2 <= end2)
	{
		if (arr[begin1] < arr[begin2])
		{
			tmp[i++] = arr[begin1++];
		}
		else if (arr[begin1]>arr[begin2])
		{
			tmp[i++] = arr[begin2++];
		}
	}

	while (begin1 <= end1)
	{
		tmp[i++] = arr[begin1++];
	}
	while (begin2 <= end2)
	{
		tmp[i++] = arr[begin2++];
	}

	for (int j = left; j <= right; j++)
	{
		arr[j] = tmp[j];
	}
}
void MergeSort(int* arr, int sz)
{
	int* tmp = new int[sz];
	_MergeSort(arr, 0, sz - 1, tmp);
	delete[] tmp;
}

归并排序非递归实现 ,借助gap值来归并:

注意归并时候发生的三种情况:

void _MergeSortNonR(int* arr, int begin1, int end1, int begin2, int end2, int* tmp)
{
	int i = begin1;
	int j = begin1;
	while (begin1 <= end1&&begin2 <= end2)
	{
		if (arr[begin1] < arr[begin2])
		{
			tmp[i++] = arr[begin1++];
		}
		else
		{
			tmp[i++] = arr[begin2++];
		}
	}
	
	while (begin1 <= end1)
	{
		tmp[i++] = arr[begin1++];
	}
	while (begin2 <= end2)
	{
		tmp[i++] = arr[begin2++];
	}

	for (; j <= end2; j++)//begin1加加j不受到影响
	{
		arr[j] = tmp[j];
	}
}
void MergeSortNonR(int* arr, int sz)
{
	int* tmp = new int[sz];

	int gap = 1;
	while (gap < sz)
	{
		for (int i = 0; i < sz; i += 2 * gap)
		{
			int begin1 = i;
			int end1 = i + gap - 1;

			int begin2 = i + gap;
			int end2 = i + 2 * gap - 1;

			//如果第二个小区间不存在了
			if (begin2 >= sz)
			{
				break;
			}

			//如果第二个小区间存在,但是第二个小区间不够gap个,结束位置就越界了,需要修正一下
			if (end2 >= sz)
			{
				end2 = sz - 1;	
			}
			_MergeSortNonR(arr, begin1, end1, begin2, end2, tmp);
		}
	}
}

三、归并排序应用场景

四、归并排序总结

相关文章