Remove Nth node from End of list--LeetCode

最新推荐文章于 2025-08-06 19:08:55 发布

原创最新推荐文章于 2025-08-06 19:08:55 发布 · 736 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c语言 #LeetCode

LeetCode 专栏收录该内容

164 篇文章

订阅专栏

本文介绍了一种高效算法，用于删除单链表中倒数第N个节点，仅通过一次遍历实现。该方法利用双指针技巧，确保了操作的简洁性和性能的优化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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.

思路：

建立两个相隔n的指针，两个指针同时向外遍历。当后一个指针到链表末尾时，删除第一个指针对应的节点。不过为了更加方便的删除是的第一个节点指向被删除节点的前一个节点最好

#include <iostream>
#include <vector> 
using namespace std;
/*
 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.
*/
/*
删除链表的倒数第n个节点 
使用两个指针来开始遍历 中间间隙可以为n，等到后一个节点达到末尾，删除前面的那个节点 
假如我们需要删除第N个节点，那么是的first和second之间的距离为n，等到 second的节点为Null,
那么我们就删除first->next节点即可 
*/

typedef struct list_node List;
struct list_node
{
	struct list_node* next;
	int value;
};

void print_list(List* list)
{
	List* tmp=list;
	while(tmp != NULL)
	{
		cout<<tmp->value<<endl;
		tmp = tmp->next; 
	}
}

/*
初始化List  将从1~n的数字插入到链表中 
*/
void Init_List(List*& head,int n)
{
	head = NULL;
	List* tmp;
	List* record;
	/*
	for(int i=1;i<=n;i++)
	{
		tmp = new List;
		tmp->next = head;
		tmp->value = i;
		head = tmp;
 
	}
	*/
	for(int i=1;i<=n;i++)
	{
		tmp = new List;
		tmp->next = NULL;
		tmp->value = i;
		if(head == NULL)
		{
			head = tmp;
			record = head;
		}
		else
		{
			record->next = tmp;
			record = tmp;
		}
	}
}

int Len_list(List* list)
{
	if(list == NULL)
		return 0;
	else
		return Len_list(list->next)+1;
}


void Remove_nth(List*&list ,int nth)
{
	List* first;
	List* second;
	int i;
	first = second = list;
	for(i=1;i<=nth && second !=NULL;i++)
		second = second->next;
	 
	while(second != NULL && second->next != NULL)
	{
		first = first->next;
		second = second->next;
	}
 
	if(first == list)
	{
		
		list = first->next;
		delete first;
	}
	else
	{
		second = first->next;
		first->next = second->next;
		delete second;	
	}
	
}

int main()
{
	List* head;
	Init_List(head,10);
	Remove_nth(head,3);
	
	print_list(head);
	return 0;
}