[LeetCode] Remove Nth Node From End of List

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签：数据结构与算法面试

原文链接：http://www.cnblogs.com/vincently/p/4060009.html

本文介绍了一种利用快慢指针的方法来解决链表中删除倒数第N个节点的问题。通过在一次遍历中实现，该方法在时间和空间复杂度上都有很好的表现。同时，文章详细阐述了代码实现过程，包括如何确保代码的鲁棒性和正确处理边界情况。

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.

思路：快慢指针。注意代码鲁棒性。时间复杂度O(n),空间复杂度O(1)

相关题目：《剑指offer》面试题15

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11 ListNode *removeNthFromEnd(ListNode *head, int n) {
12     if (head == NULL || n <= 0) return head;
13     ListNode *pfast = head;
14     for (int i = 1; i < n; ++i) {
15         if (pfast == NULL) break; //这是为了k的值大于链表节点总数的情况，本题不存在这种情况
16         pfast = pfast->next;
17     }
18     ListNode **pslow = &head;
19     while (pfast->next != NULL) {
20         pfast = pfast->next;
21         pslow = &((*pslow)->next);
22     }
23     
24     ListNode *q = *pslow;
25     *pslow = q->next;
26     delete q;
27     
28     return head;
29 }
30 };