19 删除链表的倒数第 N 个结点

题目链接： LeetCode

19 删除链表的倒数第 N 个结点

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 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

进阶：你能尝试使用一趟扫描实现吗？

Related Topics: 链表, 双指针

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解题代码

/*** @author Charlie Zhang* @version v1.0* @date 2025-06-17 05:58:04* @description 19 删除链表的倒数第 N 个结点*/
public class RemoveNthNodeFromEndOfList {public static void main(String[] args) {Solution solution = new RemoveNthNodeFromEndOfList().new Solution();}class ListNode {int val;ListNode next;ListNode() {}ListNode(int val) {this.val = val;}ListNode(int val, ListNode next) {this.val = val;this.next = next;}}/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode() {}* ListNode(int val) { this.val = val; }* ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode dummy = new ListNode(-1);dummy.next = head;ListNode slow = dummy;ListNode fast = dummy;while (n-- > 0) {fast = fast.next;}fast = fast.next; // 快指针先走1步while (fast != null) {slow = slow.next;fast = fast.next;}slow.next = slow.next.next;return dummy.next;}}}