LintCode_035_翻转链表

问题描述

翻转一个链表

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

求解

class Solution {public:    /**     * @param head: The first node of linked list.     * @return: The new head of reversed linked list.     */    ListNode *reverse(ListNode *head) {        ListNode dummy(0);        while (head) {            auto tmp = head->next;            head->next = dummy.next;            dummy.next = head;            head = tmp;        }        return dummy.next;    }};