两次遍历
思路:
先遍历一次得到数组长度length，第二次遍历找到位置在length-n的节点p，让p.next=p.next.next即可
代码：
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode: result = ListNode(0) result.next = head length = 0 first = head while first: first = first.next length +=1 length -= n first = head for i in range(length-1): first = first.next first.next = first.next.next return result.next
快慢指针
两个指针距离为n，同时移动，只需要一次遍历，当快指针到终点，慢指针对应位置，做删除节点操作。
代码：
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode: result = ListNode(0) result.next,fast,slow = head,result,result for i in range(n): fast = fast.next while fast: fast = fast.next slow = slow.next if fast else slow slow.next = slow.next.next return result.next
leetcode 每日一题 19. 删除链表的倒数第N个节点
原文：https://www.cnblogs.com/nilhxzcode/p/12812591.html