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;
}