LeetCode4：Add Two Numbers

题目：

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

解题思路：

这题相当于两个大数相加，只不过这里采用的链表的形式，而不是字符串。

解题时最需注意的是，最后一个节点要考虑会不会进位，over =1时，需要增加一个节点。

实现代码：

#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)//后一个节点，要考虑有没进位 

         {

             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)//同样，最后一个节点，要考虑有没进位 

         {

             ListNode *node = new ListNode(over);

             tnode->next = node;

         }         

         return l3->next;

        

    }



};

int main(void)

{

    return 0;

}