/*

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.

5
5
01

1
99
001

9
999
8001

*/

/*
**
 * 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* k = new ListNode(0);
        ListNode* q = k;
        int i = 0;
        while(l1 != NULL || l2 != NULL){
            int x = (l1 != NULL) ? l1->val : 0;
            int y = (l2 != NULL) ? l2->val : 0;
            int j = x + y + i;
            i = j/10;
            k->next = new ListNode(j%10);
            k = k->next;
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        } 
        if(i > 0){
            ListNode* p = new ListNode(i);
            k->next = p;
        }
        return q->next;
    }
};
*/


/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 
 //最终的优化程序 ，需要创建新的结点，C++ 创建新结点的方法和C稍有区别
 //C++用new关键字来开辟空间，C语言用 malloc() 函数开辟内存空间 
	 
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* k = new ListNode(0);
        ListNode* q = k;
        int i = 0;
        while(l1 != NULL || l2 != NULL || i){
            int j = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + i;
            i = j/10;
            k->next = new ListNode(j%10);
            k = k->next;
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        } 
        return q->next;
    }
};