[LintCode] 删除链表中倒数第n个节点

最新推荐文章于 2025-09-10 15:24:29 发布

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最新推荐文章于 2025-09-10 15:24:29 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：https://yq.aliyun.com/articles/331467

本文介绍了一种高效算法，用于从链表中删除倒数第N个节点。该方法通过使用两个指针预设固定距离，并同步移动它们来定位目标节点，从而实现仅遍历一次链表即可完成操作。

 1 /**
 2  * Definition of ListNode
 3  * class ListNode {
 4  * public:
 5  *     int val;
 6  *     ListNode *next;
 7  *     ListNode(int val) {
 8  *         this->val = val;
 9  *         this->next = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14 public:
15     /**
16      * @param head: The first node of linked list.
17      * @param n: An integer.
18      * @return: The head of linked list.
19      */
20     ListNode *removeNthFromEnd(ListNode *head, int n) {
21         // write your code here
22         ListNode* pre = head;
23         ListNode* cur = head;
24         for (int i = 0; i < n; i++)
25             cur = cur -> next;
26         if (!cur) {
27             delete pre;
28             return pre -> next;
29         }
30         while (cur -> next) {
31             pre = pre -> next;
32             cur = cur -> next;
33         }
34         delete pre -> next;
35         pre -> next = pre -> next -> next;
36         return head;
37     }
38 };