題目：

You are given two linked lists representing two non-negative numbers. 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.

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

解題思路：

這題相當於兩個大數相加，只不過這裡採用的連結串列的形式，而不是字串。

解題時最需注意的是，最後一個節點要考慮會不會進位，over =1時，需要增加一個節點。

實現程式碼：

#include <iostream>
using namespace std;

/**
You are given two linked lists representing two non-negative numbers. 
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.

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

 
*/


struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        if(l1 == NULL && l2 == NULL)
            return NULL;
         ListNode *l3 = new
 
 ListNode(-1);
         ListNode *tnode = l3;
         int over = 0;
         while(l1 && l2)
         {
             int sum = l1->val + l2->val + over;
             ListNode *node = new ListNode(sum % 10);
             over = sum / 10;
             tnode->next = node;
             tnode = tnode->next;
             l1 = l1->next;
             l2 = l2->next;
         }
         if(l1 == NULL && l2 == NULL && over)//後一個節點，要考慮有沒進位 
         {
             ListNode *node = new ListNode(over);
             tnode->next = node;
             return l3->next;
         }
             
         ListNode *left = l1;
         if(l2)
             left = l2;
         while(left)
         {
             int sum = left->val + over;
             ListNode *node = new ListNode(sum % 10);
             over = sum / 10;
             tnode->next = node;
             tnode = tnode->next;
             left = left->next;

         }
        if(over)//同樣，最後一個節點，要考慮有沒進位 
         {
             ListNode *node = new ListNode(over);
             tnode->next = node;
         }         
         return l3->next;
        
    }

};
int main(void)
{
    return 0;
}