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.

除錯了5次才成功。。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int ex,tag=0;
        ListNode *p1 = l1;
        ListNode *p2 = l2;
        ListNode *tp1,*tp2;
        while(p1!=nullptr && p2!=nullptr)
        {
            ex = p1->val+p2->val+tag;
            if(ex>9)
            {
                p1->val = ex%10;
                tag = ex/10;
            }
            else
            {
                p1->val = ex;
                tag = 0;
            }
            tp1 = p1;
            tp2 = p2;
            p1 = p1->next;
            p2 = p2->next;
        }
        while(p1!=nullptr)
        {
            ex = p1->val + tag;
            if(ex>9)
            {
                p1->val = ex%10;
                tag = ex/10;
            }
            else
            {
                p1->val = ex;
                tag = 0;
            }
            tp1 = p1;
            p1 = p1->next;
        }
        if(p1==nullptr && p2==nullptr && tag!=0)
        {
            ListNode *tmp = new ListNode(tag);
            tp1->next = tmp;
            tag = 0;
        }
        if(p2!=nullptr)
            tp1->next = p2;
        while(p2!=nullptr)
        {
            ex = p2->val + tag;
            if(ex>9)
            {
                p2->val = ex%10;
                tag = ex/10;
            }
            else
            {
                p2->val = ex;
                tag = 0;
            }
            tp2 = p2;
            p2 = p2->next;
        }
        if(p1==nullptr && p2==nullptr && tag!=0)
        {
            ListNode *tmp = new ListNode(tag);
            tp2->next = tmp;
        }
        
        return l1;
    }
};