题目描述
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
参考代码
第一次遍历获取链表的长度,第二次遍历到达要截取的位数移除节点。
# 24 ms 15 MB
# 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: ListNode, n: int) -> ListNode:
lens = 1
temp = head
while temp.next:
temp = temp.next
lens += 1
if lens < n:
return None
elif lens == n:
return head.next
else:
temp = head
for i in range(lens - n - 1):
temp = temp.next
if n == 1:
temp.next = None
else:
temp.next = temp.next.next
return head
本文摘自 :https://blog.51cto.com/u