19. Remove Nth Node From End of List

最新推荐文章于 2025-09-06 22:33:46 发布

转载最新推荐文章于 2025-09-06 22:33:46 发布 · 35 阅读

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原文链接：http://www.cnblogs.com/panini/p/6698135.html

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#数据结构与算法

题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

链接：https://leetcode.com/problems/remove-nth-node-from-end-of-list/#/description

4/11/2017

87%, 14ms

要了解每个变量的意义，想不出来就画图，不要猜初始值定在哪里
fast也可以设在dummy上，这样第8，12行就按照fast.next来判断，不过运行时间会加到18ms

 1 public class Solution {
 2     public ListNode removeNthFromEnd(ListNode head, int n) {
 3         if (head == null || n <= 0) return head;
 4         ListNode dummy = new ListNode(-1);
 5         dummy.next = head;
 6         ListNode fast = head;
 7         ListNode slow = dummy;
 8         while (n > 0 && fast != null) {
 9             fast = fast.next;
10             n--;
11         }
12         while (fast != null) {
13             fast = fast.next;
14             slow = slow.next;
15         }
16         slow.next = slow.next.next;
17         return dummy.next;
18     }
19 }