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. 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…

①英文题目 Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL ②中文题目 反转一个单链表. 示例: 输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL ③思路 用头插法(我暂时还不确定头插法是个啥,尾插法是个…

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 =…

Reverse a singly linked list. 做II之前应该先来做1的,这个倒是很简单,基本上不用考虑什么,简单的链表反转而已: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* rev…

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…

problem: Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes,…

题目内容 题目来源:LeetCode 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 conditio…

题目内容 题目来源:LeetCode Reverse a singly linked list. 题目思路 这个属于经典问题,链表反转的思路基本上已经非常固定了.有两种非常常见的方法:1.三指针法 2.头插法 这个题目用到的是三指针法. 方法:设立三个指针,分别叫做pre, curr, next.这三个指针的功能分别是:现用next保存curr的下一个结点,免得以后找不到了.然后将curr指向pre,这一步实现了反转,然后让pre指向curr, curr指向next. 具体的解释这个博客写的非常…

题目 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 ≤…