Solution { public: ListNode* reverseList(ListNode* head) { if (head == NULL || head->next == NULL){ return head; } ListNode* newhead = reverseList(head->next); head->next->next = head; head->next = NULL; return
Reverse a singly linked list. 反转链表,我这里是采用头插法来实现反转链表。 代码如下: /*class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } */ public class Solution { public ListNode reverseList(ListNode head) { if(head==null || head.next==null) return head; L...
AI代码解释 publicclassSolution{publicListNodereverseBetween(ListNode head,int m,int n){if(m==n||head==null||head.next==null){returnhead;}ListNode pre=newListNode(0);pre.next=head;ListNode Mpre=pre;ListNode NodeM=head;ListNode NodeN=head;int mNum=1;int nNum=1;while(mNum<m&&NodeM!=...
题目链接:https://leetcode.com/problems/reverse-linked-list/ 方法一:迭代反转 https://blog.csdn.net/qq_17550379/article/details/80647926讲的很清楚 方法二:递归反转 解决递归问题从最简单的c
https://leetcode.com/problems/reverse-linked-list/ 题目: Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 思路: 头插法建立链表 算法: 1. public ListNode reverseList(ListNode head) { ...
问题链接英文网站:92. Reverse Linked List II中文网站:92. 反转链表 II问题描述Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the l…
方法一、迭代法 # Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x# self.next = NoneclassSolution(object):defreverseList(self,head):""" :type head: ListNode :rtype: ListNode
LeetCode 206. 反转链表(Reverse Linked List) 示例: 输入:1->2->3->4->5->NULL输出:5->4->3->2->1->NULL 切题 一、Clarification 只需注意为空链表的情况 二、Possible Solution 1、迭代 2、递归 可利用哨兵简化实现难度 Python3 实现
[LeetCode]92.Reverse Linked List II 【题目】 Reverse a linked list from positionmton. Do it in-place and in one-pass. For example: Given1->2->3->4->5->NULL,m= 2 andn= 4, return1->4->3->2->5->NULL. Note: Givenm,nsatisfy the following condition:...
【摘要】 这是一道关于链表翻转的LeetCode题目,希望对您有所帮助。 题目概述: Reverse a singly linked list. 翻转一个单链表,如:1->2 输出 2->1;1->2->3 输出3->2->1。 题目解析: 本人真的比较笨啊!首先想到的方法就是通过判断链尾是否存在,再新建一个链表,每次移动head的链尾元素,并删除head...