题解 | #反转链表#

解法一：三个指针beg，mid，end。每次mid->next指向beg，然后三个指针都向后移动一位，知道end==null。最后一步是mid->next = beg。返回mid

<?php

/*class ListNode{
    var $val;
    var $next = NULL;
    function __construct($x){
        $this->val = $x;
    }
}*/

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @return ListNode类
 */
function ReverseList($head)
{
    // write code here
    if($head->next == null){
        return $head;
    }
    $beg = null;
    $mid = $head;
    $end = $head->next;
    while($end != null){
        $mid->next = $beg;
        $beg = $mid;
        $mid = $end;
        $end = $mid->next;
    }
    $mid->next = $beg;
    return $mid;
}