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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 * 类似大整数加法的实现
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

        int carry = 0;
        ListNode* res = new ListNode(0);
        ListNode* it = res;

        while(l1!=NULL || l2!=NULL)
        {
            int num1 = 0, num2 = 0;
            if(l1 != NULL)
            {
                num1 = l1->val;
                l1 = l1->next;
            }

            if(l2 != NULL){
                num2 = l2->val;
                l2 = l2->next;
            }

            int t = num1 + num2 + carry;
            it->next = new ListNode(t % 10);
            carry = t / 10;//进位
            it = it->next;
        }

        while(carry > 0)
        {
            it->next = new ListNode(carry % 10);
            carry /= 10;
        }
        return res->next;

    }
};