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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

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**题目大意:**342+465=807两个数相加，每个数都是以倒叙排列在链表中。
分析：链表的使用，如何向链表添加元素，以及进位的处理。
代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {
            val = x;
            next = NULL;
 }
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode *answer = new ListNode(0);//新建一个头结点
        ListNode *newone;//用来当作链表的指针用
        newone = answer;
        int sum = 0,temp = 0;
        while(l1!=NULL && l2 != NULL)
        {
            sum = sum + l1->val + l2->val;
            temp = sum %10;
            sum = sum/10;
            newone->next = new ListNode(temp);
            newone = newone->next;

            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1!=NULL)
        {
            sum = sum + l1->val;
            temp = sum %10;
            sum = sum/10;
            newone->next = new ListNode(temp);
            newone = newone->next;

            l1 = l1->next;
        }
        while(l2 != NULL)
        {
            sum = sum + l2->val;
            temp = sum %10;
            sum = sum/10;
            newone->next = new ListNode(temp);
            newone = newone->next;

            l2 = l2->next;
        }
        if(sum !=0 )
        {
            newone->next = new ListNode(sum);
            newone = newone->next;
        }
        return answer->next;
    }
};

也有更为简单的写法，但目前就可以掌握这个，正在学习其他的写法，如有不对，敬请指教