1.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, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: 1 ? m ? n ? length of list. 2.解...
代码: 1/**2* Definition for singly-linked list.3* struct ListNode {4* int val;5* ListNode *next;6* ListNode(int x) : val(x), next(NULL) {}7* };8*/9classSolution {10public:11ListNode *reverseBetween(ListNode *head,intm,intn) {12if(!head)returnNULL;13ListNode *pre =NULL;14L...
next return head ## 头元素指代一个链表 ## 将链表转化为数组,输出 def ll_to_list(head): cur = head list = [] while cur: list.append(cur.val) cur=cur.next return list class Solution: def reverseList(self, head): """ :type head: ListNode 输入链表节点 :rtype: ListNode 返回链表节点...
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!=nu...
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) ...
Reverse a singly linked list. 翻转一个链表 #1 第一种方法:迭代 代码语言:javascript 代码运行次数:0 classListNode(object):def__init__(self,x):self.val=x self.next=NoneclassSolution(object):defreverseList(self,head):""":type head:ListNode:rtype:ListNode""" ...
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, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: ...
Reverse Linked List 思路:从头开始第一个节点是q,而p指向q的下一节点,用t存p->next,然后让p指向q,然后更新q=p,p指向其下一个。 ...LeetCode 206. Reverse Linked List(链表) 题目来源:https://leetcode.com/problems/reverse-linked-list/ 问题描述 206. Reverse Linked List Easy Reverse a singly...
head || !head->next) return head; //将后面子链表反转后的尾保存一下,其实不存也可以,只是这样的逻辑更清楚 ListNode *tail = head->next; //后面子链表反转后的头就是整个链表反转后的头 ListNode *newhead = reverseList(head->next); //将后面子链表的尾连到第一个节点上 tail->next = head; ...
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,