/**
* 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) {
}
};
剑指 Offer 24. 反转链表
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
注意：本题与主站 206 题相同：https://leetcode.cn/problems/reverse-linked-list/
原站题解
golang 解法, 执行用时: 0 ms, 内存消耗: 2.5 MB, 提交时间: 2021-05-28 11:47:48
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
cur := head
var prev *ListNode
var next *ListNode
for cur != nil {
next = cur.Next
cur.Next = prev
prev = cur
cur = next
}
return prev
}
php 解法, 执行用时: 4 ms, 内存消耗: 16.8 MB, 提交时间: 2021-05-28 11:45:11
/**
* Definition for a singly-linked list.
* class ListNode {
* public $val = 0;
* public $next = null;
* function __construct($val) { $this->val = $val; }
* }
*/
class Solution {
/**
* @param ListNode $head
* @return ListNode
*/
function reverseList($head) {
$cur = $head;
$next = null;
$prev = null;
while ( $cur != null ) {
$next = $cur->next;
$cur->next = $prev;
// $cur->prev = $next;
$prev = $cur;
$cur = $next;
}
return $prev;
}
}