[转载请注明]http://www.cnblogs.com/igoslly/p/8670038.html 链表逆序在链表题目中还是较为常见的,这里将Leetcode中的两道题放在一起,分别是 0092 Reverse Linked List  II 和 0206 Reverse Linked List,0092在0206的基础上,提出了部分逆序的概念 首先来看0206,题目描述为:Reverse a singly linked list. 也就是给出一个链表,要求我们将链表全部倒置链接 想法1:…

Reverse a singly linked list. Example:           Input: 1->2->3->4->5->NULL                   Output: 5->4->3->2->1->NULL Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? 解…

Reverse a singly linked list. click to show more hints. Subscribe to see which companies asked this question. 利用循环. public class Solution { public ListNode reverseList(ListNode head) { if(head == null ||head.next == null) return head; ListNode pre =…

将单向链表反转 完成如图操作,依次进行即可 1 2 3 /** * 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) { ListNode* now = h…

链表反转,一发成功~ /** * 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) { //就地反转 if(head == NULL || head ->n…

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? 之前做到Reverse Linked List II 倒置链表之二的时候我还纳闷怎么只有二没有一呢,原来真是忘了啊,现在才加上,这道题跟之前那道比起来简单不少,题目为了增加些许难度,让我们分别用…

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 ≤ lengt…

1. Reverse Linked List 题目链接 题目要求: Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 初想用递归来实现应该会挺好的,但最终运行时间有点久,达到72ms,虽然没超时.具体程序如下: /** * Definition for singly-linked list. *…

2018-09-11 22:58:29 一.Reverse Linked List 问题描述: 问题求解: 解法一:Iteratively,不断执行插入操作. public ListNode reverseList(ListNode head) { if (head == null) return null; ListNode dummy = new ListNode(-1); dummy.next = head; ListNode cur = head; ListNode then = nul…

表不支持随机查找,通常是使用next指针进行操作. 206. 反转链表 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ //时间:O(n) 只遍历了一遍链表 //空间:O(1) 开了三个指针的空间 class Solution { public: ListNode*…