19. 删除链表的倒数第 N 个结点
Difficulty: 中等
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。
进阶:你能尝试使用一趟扫描实现吗?
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
Solution
快慢指针方法,让快指针先走n步,然后快指针和慢指针同时走,此时快指针的距离为n,当快指针将要到达链表末尾的时候,慢指针的下一个节点即为需要删除的节点。思路挺巧妙的,有点意思。
# 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:
if not head:
return None
slow, fast = head, head
for _ in range(n):
fast = fast.next
# 删除第一个节点
if not fast:
return head.next
else:
while fast.next:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head