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...
Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list. Example 1: Input: head = [1,2,3,4,5], left = 2, right = 4 Output: [1,4,3,2,5] ...
1/**2* Definition for singly-linked list.3* public class ListNode {4* int val;5* ListNode next;6* ListNode(int x) {7* val = x;8* next = null;9* }10* }11*/12publicclassSolution {13publicListNode reverseBetween(ListNode head,intm,intn) {14ListNode prev =newListNode(-1);15prev...
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: 1≤m≤n≤ length of list. 把[m,n]那一段抠出来,reverse之后,再拼回...
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, ...
public class Solution { public ListNode reverseBetween(ListNode head, int m, int n) { if(head==null || head.next==null) return head; ListNode newHead=new ListNode(0); newHead.next=head; ListNode pre=newHead,start=null,then=null; ...
LeetCode-链表 链表(Linked List)是一种常见的基础数据结构,是一种线性表,但是并不会按线性的顺... raincoffee阅读 1,218评论 0赞 6 Linked List的复习总结 Single Linked List 相比较另一个基本的数据结构array,linked list有几个优势:尺寸... dol_re_mi阅读 8,197评论 0赞 3 穿越到金庸武侠世界中的囧...
leetcode206题解: https://www.jianshu.com/p/e3b1ed444819 加入了测试数据 C++代码: #include<algorithm>#include<iostream>#include<queue>usingnamespacestd;structListNode{intval;ListNode*next;ListNode(intx):val(x),next(NULL){}};classSolution{public:ListNode*begin;ListNode*end;ListNode*reverseBetween(...
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? 既然问了能否iteratively or recursively, 那就both把. iterative 解法: 总结就是得到下一个节点,更改当前节点指向,将指针往下移动,直到过完整个linked...
Fork= 3, you should return:3->2->1->4->5 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode reverseKGroup(ListNode head, int k) { ...