题目:
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:
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
解答:
python
# 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,使其下一个节点为头节点
dummy_head = ListNode(0,head)
# 设置一个快指针和一个慢指针均指向虚拟节点
fast = slow = dummy_head
# 使fast指针先走n+1步
for i in range(n+1):
fast = fast.next
# 两个指针一起向前,当fast指针指向null时,跳出循环,此时slow指针指向倒数n+1节点
while fast:
fast = fast.next
slow = slow.next
# 改变slow指针关于下一个节点的指向
slow.next = slow.next.next
return dummy_head.next