2-Add Two Numbers

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

题意：就是用编程模拟笔算

注意：两个数的位数可能不同，所以可能出现的情况是当一个数加完了最高位，另一个数还未到最高位。所以只要有一个没达到我们就累加。

     循环条件，数字没处理完或处理完了还有进位

难点：怎么建立两个结点之间的关系

思路：使用游标指向当前正在操作结点的前一个结点，首结点前面是NULL，如果我们操作的是首结点，结束后移动游标。否则我们申请一个节点赋值，建立关系并且移动游标。

/**
 * 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) {
        
        ListNode* head = new ListNode(0);
        ListNode* last = NULL;
        int sum = 0;
        
        while( (l1 != NULL ) || ( l2 != NULL ) || ( sum != 0 ) )
        {
            if( l1 != NULL )
            {
                sum += l1->val;
                l1 = l1->next;
            }

            if( l2 != NULL )
            {
                sum += l2->val;
                l2 = l2->next;
            }    
            
            if( last != NULL )
            {
                ListNode* t = new ListNode(sum % 10);
                last->next = t;
                last = last->next;
            }
            else
            {
                head->val = sum % 10;
                last = head;
            }
            
            sum /= 10;
        }
        
        return head;
    }
};