给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。

请你将两个数相加,并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外,这两个数都不会以 0 开头。

 

示例 1:

输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807

示例 2:

输入:l1 = [0], l2 = [0]
输出:[0]

示例 3:

输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1

提示:

每个链表中的节点数在范围 [1, 100] 内
0 <= Node.val <= 9
题目数据保证列表表示的数字不含前导零

 

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/add-two-numbers
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

 

自解-解法如下

/**
 * Definition for a singly-linked list.
 */
class ListNode {
     public $val = 0;
     public $next = null;
     function __construct($val = 0, $next = null) {
         $this->val = $val;
         $this->next = $next;
     }
}

class Solution {

    /**
     * @param ListNode $l1
     * @param ListNode $l2
     * @return ListNode
     */
    function addTwoNumbers($l1, $l2) {
        $ret = 0;
        $hex = 10;

        $cur1Node = $l1;
        $cur2Node = $l2;
        for ($i = 0; ; $i++) {
            $curNum1 = $cur1Node->val?? 0;
            $curNum2 = $cur2Node->val?? 0;

            $sumTemp = $curNum1 + $curNum2;

            $sumTemp = $sumTemp * square($hex, $i);

            $ret += $sumTemp;   

            $cur1Node = $cur1Node->next?? null;
            $cur2Node = $cur2Node->next?? null;
            if (null == $cur1Node && null == $cur2Node) {
                break;
            }
        }
        echo $ret;
        echo "\n";
    }
}

function square($num, $multiplication) 
{
    if ($multiplication == 0) {
        return 1;
    }

    if ($multiplication == 1) {
        return $num;
    }

    for ($i = 1; $i < $multiplication; $i++) {
        $num *= $num;
    }
    return $num;
}


$list1 = new ListNode(3, null);
$list1->next = new ListNode(3, null);
$list1->next->next = new ListNode(4, null);



$list2 = new ListNode(8, null);
$list2->next = new ListNode(7, null);
$list2->next->next = new ListNode(3, null);


$obj = new Solution();
$obj->addTwoNumbers($list1, $list2);

 

posted on 2021-08-10 11:30  黑熊一只  阅读(136)  评论(0)    收藏  举报