題目：

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.

題意：

兩連結串列從尾到頭組成的數字的和

思路：

一開始直接求兩個數的值，然後a+b=c，把c的值構成一個連結串列，WA了。我以為是int的範圍太小改成long long，還是不行才發現原來是大數加法~難受啊，沒法子只能推倒重來~其實還是蠻簡單的，只不過要細心細心再細心！！！

Code：

/**
 * 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 sum=0,flag=0;
        ListNode *l3,*head;
        while(l1&&l2){
            sum=sum+l1->val+l2->val;
            if(flag==0){
                l3=new ListNode(sum%10);
                head=l3;
                flag=1;
            }
            else{
                head->next=new ListNode(sum%10);
                head=head->next;
            }
            sum/=10;
            l1=l1->next;
            l2=l2->next;
        }
        while(l1){
            sum=sum+l1->val;
            head->next=new ListNode(sum%10);
            head=head->next;
            sum/=10;
            if(sum==0){
                head->next=l1->next;
                break;
            }
            l1=l1->next;
        }
        while(l2){
            sum=sum+l2->val;
            head->next=new ListNode(sum%10);
            head=head->next;
            sum/=10;
            if(sum==0){
                head->next=l2->next;
                break;
            }
            l2=l2->next;
        }
        if(sum) head->next=new ListNode(sum%10);
        return l3;
    }
};