lintcode 翻转链表

问题描述

http://www.lintcode.com/zh-cn/problem/reverse-linked-list/

样例
给出一个链表1->2->3->null，这个翻转后的链表为3->2->1->null

笔记

就是倒腾一下几个节点，找递推规律

代码

/** * Definition of ListNode *  * class ListNode { * public: *     int val; *     ListNode *next; *  *     ListNode(int val) { *         this->val = val; *         this->next = NULL; *     } * } */class Solution {public:    /**     * @param head: The first node of linked list.     * @return: The new head of reversed linked list.     */    ListNode *reverse(ListNode *head) {        // write your code here        if (head == NULL)            return NULL;        ListNode *pre = NULL;        ListNode *tmp = head->next;        while (head)        {            tmp = head->next;            head->next = pre;            pre = head;            head = tmp;        }        return pre;    }};

二次练习

/** * Definition of ListNode *  * class ListNode { * public: *     int val; *     ListNode *next; *  *     ListNode(int val) { *         this->val = val; *         this->next = NULL; *     } * } */class Solution {public:    /**     * @param head: The first node of linked list.     * @return: The new head of reversed linked list.     */    ListNode *reverse(ListNode *head) {        // write your code here        ListNode* pre = NULL;        ListNode* now = head;        ListNode* nex = NULL;        while (now != NULL)        {            nex = now->next;            now->next = pre;            pre = now;            now = nex;        }        return pre;    }};