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...
* @param head: The first node of linked list. * @return: The new head of reversed linked list.*/ListNode*reverse(ListNode *head) {//case1: empty listif(head == NULL)returnhead;//case2: only one element listif(head->next == NULL)returnhead;//case3: reverse from the rest after ...
* * Hint: * A linked list can be reversed either iteratively or recursively. Could you implement both? */ public class ReverseLinkedList { /** * * 反转单向链表 * * 使用迭代的方法 * * @param head * @return */ public Node reverseByIterative (Node head) { Node pre = null; while (...
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,
Reverse Linked List I 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? 递归法 复杂度 时间O(N) 空间 O(N) 递归栈空间 思路 基本递归 ...
代码例如以下: AI检测代码解析 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseList(ListNode* head) { ...
Reverse Linked List II 题目大意 翻转指定位置的链表 解题思路 将中间的执行翻转,再将前后接上 代码 迭代 AI检测代码解析 class Solution(object): # 迭代 def reverseBetween(self, head, m, n): """ :type head: ListNode :type m: int :type n: int ...
反转链表(Reverse Linked List) 反转一个单链表。如下示例:: 输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL public class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } 一、 迭代法: 注意观察示例:1->2->3->4->5->NULL的反转可以看成:NULL<-1<-...
Loading...leetcode.com/problems/reverse-linked-list/ 题目很简单:输入一个单链表的头节点,返回反转后的链表的头节点。例如原始的链表是:1 -> 2 -> 3 -> 4 -> 5 -> null(或者None), 反转后的链表是: 5 -> 4 -> 3 -> 2 -> 1 -> null(或者None) ...
(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]...