Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

原创最新推荐文章于 2024-01-12 15:01:11 发布 · 1.3k 阅读

0 ·

CC 4.0 BY-SA版权

Leedcode 专栏收录该内容

72 篇文章

订阅专栏

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。通过使用双指针技巧，即一个快指针和一个慢指针，能够在遍历一次链表的过程中找到并删除目标节点。

Given a linked list, remove the nthnode 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：

两个指针，p先走n步，然后q开始走。当p走到末尾时，q正好到了待删除节点。

#include <iostream>
using namespace std;
struct LinkNode
{
	int data;
	LinkNode *next;
	LinkNode(int x):data(x),next(NULL){}
};

LinkNode * removeNthNodeFromEnd(LinkNode *head,int n)
{
	if(head == NULL || n == 0)//防御性编程，增加健壮性
		return head;
	LinkNode *dummy = new LinkNode(-1);
	dummy->next = head;
	LinkNode *p,*q,*tmp;
	p = dummy;
	q = dummy;
	
	//p先走n步
	for(int i=0;i<n;i++)
	{
		if(p->next != NULL)
			p = p->next;
		else
			return head;//如果n超过链表的长度，直接返回head
	}

	//然后p,q一起走
	while(p->next)
	{
		p = p->next;
		q = q->next;
	}

	//此时q指向的是待删除节点的前驱
	tmp = q->next->next;
	delete q->next;
	q->next = tmp;

	head = dummy->next;
	delete dummy;
	return head;
}