LeetCode(LinkedList)Add Two Numbers

Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

思路：一开始想到了借助字符串，后来发现在转型过程中容易出错，遂弃用。看到了一个人的解法还不错，思路比较清晰，就拿过来自己重写了一下。代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null){
            return l2;
        }
        if(l2==null){
            return l1;
        }
        //用于存储后续结点
        ListNode head=new ListNode(0);
        ListNode p=head;
        int temp=0; //初始进位为0
        //当满足l1,l2任意不为空，或者仍然有进位时进入循环
        while(l1!=null || l2!=null || temp!=0){
            if(l1!=null){
                temp+=l1.val;
                l1=l1.next;
            }
            if(l2!=null){
                temp+=l2.val;
                l2=l2.next;
            }
            p.next=new ListNode(temp%10);
            p=p.next;
            temp/=10;
        }
        return head.next;
    }
}