剑指 Offer 22. 链表中倒数第k个节点 (Python 实现)

x33g5p2x  于2022-06-29 转载在 Python  
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题目:
输入一个链表,输出该链表中倒数第k个节点。为了符合大多数人的习惯,本题从1开始计数,即链表的尾节点是倒数第1个节点。
例如,一个链表有 6 个节点,从头节点开始,它们的值依次是 1、2、3、4、5、6。这个链表的倒数第 3 个节点是值为 4 的节点。

示例:

给定一个链表: 1->2->3->4->5, 和 k = 2.
返回链表 4->5.

code:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        temp = head
        n = 0
        while temp:
            n+=1
            temp = temp.next
        for _ in range(n-k):
            head = head.next
        return head
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        fast, slow = head, head
        for _ in range(k-1):
            fast = fast.next
        while fast.next != None:
            fast = fast.next
            slow = slow.next
        return slow

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