[LeetCode]2. 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
代码：

/**
 * 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) {
         if(l1==NULL) return l2;
         if(l2==NULL) return l1;

         ListNode *resultHead = NULL ,*beforeNode = NULL;
         int carry = 0;//当前进位 
         int  a,b ;
         bool flag = l1 || l2;//只有两个都为NULL时，才有可能结束
         int currentResult = 0;
         while(flag||carry){//当进位为0且l1,l2都为NULL时，才结束
             ListNode *resultNode = new ListNode(0);
             if(resultHead == NULL){
                  resultHead = resultNode;
                  beforeNode = resultHead;
             }
             else{
                 beforeNode->next = resultNode;
                 beforeNode = resultNode;
             }
             a = l1 ? (l1->val) : 0;//当l为空时，将其内容当成0
             b = l2 ? (l2->val) : 0;
             currentResult = a + b + carry;
             resultNode->val = currentResult % 10;
             carry = currentResult / 10;
             //当l为NULL时，没有next，一定要让其保持为NULL
             l1 = l1 ? l1->next:l1;
             l2 = l2 ? l2->next:l2;
             flag = l1 || l2;

         }

         return resultHead;
    }
};

注：由于两个列表可能长度不同，所以这里采用了补齐的方法，将短列表后面的内容都当成0。