目录
- 1.题目
- 2.答案
- 3.提交结果截图
链接: 删除链表的倒数第 N 个结点
1.题目
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?
2.答案
/*** 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) {List<ListNode> list = new ArrayList<>();ListNode node = head;while (node != null) {list.add(node);node = node.next;}int position = list.size() - 1 - (n - 1);boolean hasBefore = position - 1 >= 0;boolean hasAfter = position + 1 <= list.size() - 1;if (hasBefore && hasAfter) {list.get(position - 1).next = list.get(position + 1);} else if (hasBefore) {list.get(position - 1).next = null;} else if (hasAfter) {head = list.get(position + 1);} else {head = null;}return head;}
}
3.提交结果截图
整理完毕,完结撒花~ 🌻