定义一个函数,输入一个链表的头节点,反转该链表并输出反转后链表的头节点。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
限制:
0 <= 节点个数 <= 5000
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/fan-zhuan-lian-biao-lcof
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遍历,同时把该节点的值移到答案前面
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode reverseList(ListNode head) { ListNode prev = null; ListNode curr = head; while (curr != null) { ListNode next = curr.next; curr.next = prev; prev = curr; curr = next; } return prev; } }
标签:24,---,ListNode,next,链表,力扣,curr,prev From: https://www.cnblogs.com/allWu/p/17231962.html