题目要求:
给你单链表的头节点 head ,请你反转链表,并返回反转后的链表。
示例 1:输入:head = [1,2,3,4,5]
输出:[5,4,3,2,1]
示例 2:输入:head = [1,2]
输出:[2,1]
示例 3:输入:head = []
输出:[]
提示:链表中节点的数目范围是 [0, 5000]
-5000 <= Node.val <= 5000
进阶:链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题?来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-linked-list
解题代码:
//Definition for singly-linked list. struct ListNode { int val; ListNode *next; ListNode() : val(0), next(nullptr) {} ListNode(int x) : val(x), next(nullptr) {} ListNode(int x, ListNode *next) : val(x), next(next) {} }; class Solution { public: ListNode *reverseList(ListNode *head) { ListNode* pre = NULL; ListNode* cur = head; while (cur != NULL) { ListNode* next = cur->next; cur->next = pre; pre = cur; cur = next; } return pre; } };