【leetcode】19. 删除链表的倒数第N个节点

题目

19. 删除链表的倒数第N个节点

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 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

代码

1. 获取长度

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):def removeNthFromEnd(self, head, n):""":type head: Optional[ListNode]:type n: int:rtype: Optional[ListNode]"""if head is None:return Nonecur = headlenth = 0while cur:cur = cur.nextlenth += 1if lenth == n:return head.nextcur = headfor i in range(lenth-n-1):cur = cur.nextcur.next = cur.next.nextreturn head

2. 双指针

思路： 快指针先走n+1步，等到NULL时，慢指针到倒数第n个的前面一个节点

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):def removeNthFromEnd(self, head, n):""":type head: Optional[ListNode]:type n: int:rtype: Optional[ListNode]"""# 双指针dummy_head = ListNode(0, head)slow = fast = dummy_headfor i in range(n+1):fast = fast.nextwhile fast:slow = slow.nextfast = fast.nextslow.next = slow.next.nextreturn dummy_head.next