LeetCode 19： Remove Nth Node From End of List

「已注销」

于 2015-06-02 22:03:47 发布

阅读量1.5k

点赞数 1

CC 4.0 BY-SA版权

分类专栏： LeetCode LeetCode算法分析文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/sunao2002002/article/details/46335987

LeetCode 同时被 2 个专栏收录

80 篇文章

订阅专栏

LeetCode算法分析

76 篇文章

订阅专栏

本文介绍了一种在单次遍历中删除链表倒数第N个节点的方法，通过双指针技巧实现。适用于链表操作的学习者。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a linked list, remove the n^th 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.

代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
                       if (head == NULL)
				return head;
			ListNode *p=head, *q = head;
			for (int i =0; i<n; i++)
			{
				if (p)
				{
					p = p->next;
				}else
					return head;
			}
			if (p == NULL)
			{
				head = head->next;
				return head;
			}
			while(p->next != NULL){
				p = p->next;
				q = q->next;
			}
			q->next = q->next->next;
			return head;
        
    }
};