python-leetcode-删除链表的倒数第 N 个结点

19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode）

# 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]:# 创建一个虚拟节点，指向头节点dummy = ListNode(0, head)first = dummysecond = dummy# 让第一个指针先走 n+1 步for _ in range(n + 1):first = first.next# 同时移动两个指针，直到第一个指针到达链表末尾while first:first = first.nextsecond = second.next# 删除倒数第 n 个节点second.next = second.next.next# 返回头节点return dummy.next