http://www.jiuzhang.com/solutions/reverse-linked-list/网页标题:Lintcode35ReverseLinkedListsolution题解 URL网址:http://www.xiwangwangguoyuan.com/article/iihgpc.html 其他资讯sqlite库go语言的简单介绍 go语言的匿名函数 go 匿名字段 域名怎么加密码
1publicclassSolution {2publicListNode reverseBetween(ListNode head,intm,intn) {3if(m >= n || head ==null) {4returnhead;5}67ListNode dummy =newListNode(0);8dummy.next =head;9head =dummy;1011for(inti = 1; i < m; i++) {12if(head ==null) {13returnnull;14}15head =head.next;...
Reverse a linked list from position m to n. Notice:Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list. 翻转链表中第m个节点到第n个节点的部分 注意:m,n满足1 ≤ m ≤ n ≤ 链表长度 【题目链接】 http://www.lintcode.com/en/problem/reverse-linked-list-ii/ 【...
Lintcode35 Reverse Linked List solution 题解 【题目描述】 Reverse a linked list. 翻转一个链表 【题目链接】 http://www.lintcode.com/en/problem/reverse-linked-list/ 【题目解析】 这题要求我们翻转[m, n]区间之间的链表。对于链表翻转来说,几乎都是通用的做法,譬如p1 -> p2 -> p3 -> p4,如果我...
// 206. Reverse Linked List // https://leetcode.com/problems/reverse-linked-list/description/ // // 递归的方式反转链表 // 时间复杂度: O(n) // 空间复杂度: O(1) class Solution { public: ListNode* reverseList(ListNode* head) { // 递归终止条件 if(head == NULL || head->next ==...
* }*/classSolution {public:/** * @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;//cas...
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!=null){Mpre=...
The Java Program to reverse a Linked List iteratively and printing its elements is given below: package com.journaldev.linkedlist.reverse; import com.journaldev.linkedlist.reverse.MyLinkedList.Node; public class ReverseLinkedList { public static void main(String[] args) { ...
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode reverseKGroup(ListNode head, int k) { if (head == null || head.next == null) return head; int count =...
http://www.lintcode.com/en/problem/reverse-linked-list-ii/ 【题目解析】 反转整个链表的变种,指定了起点和终点。由于m=1时会变动头节点,所以加入一个dummy头节点 1. 找到原链表中第m-1个节点start:反转后的部分将接回改节点后。 2. 将从p = start->next开始,长度为L = n-m+1的部分链表反转。