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** 同时也要记录第m个节点
(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 class ListNode: 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 def linkedlist(li...
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) { 2. ListNode p = head, q, t...
A linked list can be reversed either iteratively or recursively. Could you implement both? Subscribeto see which companies asked this question 解法1:一个最简单的办法就是借助栈的后进先出功能,先扫描一遍链表保存每个节点的值,然后再从头到尾遍历,将栈中元素值一一赋给链表节点。时空复杂度都是O(n)。
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
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...
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…
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,
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