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19.删除链表的倒数第N个结点 ​

快慢指针，https://leetcode.cn/problems/remove-nth-node-from-end-of-list/

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

示例 1：

img
输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

提示：

  • 链表中结点的数目为 sz
  • 1 <= sz <= 30
  • 0 <= Node.val <= 100
  • 1 <= n <= sz
python
from typing import Optional
#Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        # 创建哑结点（dummy）以处理头结点可能被删除的情况
        dummy = ListNode(0)
        dummy.next = head
        fast = slow = dummy

        # 快指针先前进 n+1 步
        for _ in range(n + 1):
            fast = fast.next

        # 快慢指针同时移动直到快指针到达链表末尾
        while fast:
            fast = fast.next
            slow = slow.next

        # 此时慢指针的下一个结点是要删除的结点
        slow.next = slow.next.next

        # 返回头结点
        return dummy.next

# 测试用例
def print_list(head):
    result = []
    while head:
        result.append(head.val)
        head = head.next
    return result

if __name__ == '__main__':
    head = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))
    sol = Solution()
    print(print_list(sol.removeNthFromEnd(head, 2)))  # 输出：[1, 2, 3, 5]

    head = ListNode(1)
    print(print_list(sol.removeNthFromEnd(head, 1)))  # 输出：[]

    head = ListNode(1, ListNode(2))
    print(print_list(sol.removeNthFromEnd(head, 1)))  # 输出：[1]