leetcode-19 Remove Nth Node From End of List

最新推荐文章于 2023-12-11 10:06:36 发布

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本文详细解析了如何在一过链表中找到并删除倒数第n个节点的算法，通过双指针法实现单次遍历完成任务。

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问题描述：

Given a linkedlist, remove the n^th node fromthe 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.

问题分析：查找链表倒数第k个节点，以及删除链表节点的知识相结合

代码：

Java解法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
		
		ListNode firstNode = head;//双指针法
		ListNode lastNode = head;
		ListNode result = head;//记录头结点返回
		ListNode preNode = null;//由于要执行删除操作，由于是单链表，故要事先记录其前驱
		
		/*寻找链表倒数第k个节点*/
		for(int i = 0; i < n; i++)
		{
			if(firstNode == null)
				return null;
			firstNode = firstNode.next;
		}
		
		while(firstNode != null)
		{
			firstNode = firstNode.next;
			preNode = lastNode;
			lastNode = lastNode.next;
		}
		
		//删除倒数第n个节点
		if(preNode == null)//删除时要注意preNode为null即lastNode为头节点的情况，注意删除头结点，则返回的头结点应该为lsatNode.next
		{
			result = head.next;
		}
		else
		{
			preNode.next = lastNode.next;
		}			
        return result;
    }
}

C++解法（基本相同）

/**
 * 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) {
        
		ListNode* firstNode = head;
		ListNode* lastNode = head;
		ListNode* result = head;
		ListNode* preNode = nullptr;
		
		for(int i = 0; i < n; i++)
		{
			if(firstNode == nullptr)
				return nullptr;
			firstNode = firstNode->next;
		}
		
		while(firstNode != nullptr)
		{
			firstNode = firstNode ->next;
			preNode = lastNode;
			lastNode = lastNode ->next;
		}
		
		if(preNode == nullptr)
		{
			result = head->next;
		}
		else
		{
			preNode-> next = lastNode-> next;
		}
		return result;
    }
};