请使用一趟扫描完成反转。 说明: 1 ≤ m ≤ n ≤ 链表长度。 示例: 输入: 1->2->3->4->5->NULL, m = 2, n = 4 输出: 1->4->3->2->5->NULL 分析 给定初始链表为 1->2->3->4->5->NULL,如图 初始状态 我们需要找到第m个节点和第n个节点,分别记为MNode和 ** NNode** 同时也要...
/** * Source : https://leetcode.com/problems/reverse-linked-list/ * * * Reverse a singly linked list. * * click to show more hints. * * Hint: * A linked list can be reversed either iteratively or recursively. Could you implement both? */ public class ReverseLinkedList { /** * ...
Can you solve this real interview question? Reverse Linked List - Given the head of a singly linked list, reverse the list, and return the reversed list. Example 1: [https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg] Input: head = [1,2,3,
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
Leetcode260反转链表(java/c++/python) JAVA: class 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 newHead; } } C++: class Solution...
leetcode 206. Reverse Linked List 反转字符串,Reverseasinglylinkedlist.反转链表,我这里是采用头插法来实现反转链表。代码如下:/*classListNode{intval;ListNodenext;ListNode(intx){val=x;}}*/publicclassSolution{publicListNoderever
LeetCode: 92. Reverse Linked List II 题目描述 Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, ...
给你单链表的头节点head,请你反转链表,并返回反转后的链表。 示例1: 输入:head = [1,2,3,4,5]输出:[5,4,3,2,1] 示例2: 输入:head = [1,2]输出:[2,1] 示例3: 输入:head = []输出:[] 提示: 链表中节点的数目范围是[0, 5000] ...
英文网站:92. Reverse Linked List II 中文网站:92. 反转链表 II 问题描述 Given theheadof a singly linked list and two integersleftandrightwhereleft <= right, reverse the nodes of the list from positionleftto positionright, and returnthe reversed list. ...
(参考视频讲解:Leetcode力扣|206反转链表|递归|reverse linked list_哔哩哔哩_bilibili) # 定义一个链表节点类classListNode:def__init__(self,val=0,next=None):# 初始化函数self.val=val# 节点的值self.next=next# 指向下一个节点的指针# 将给出的数组转换为链表deflinkedlist(list):head=ListNode(list[0]...