Leetcode #2. Add Two Numbers

 Leetcode #2. Add Two Numbers

 ◎ 難度： Medium

 ◎ 題目：

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 contains a single digit. 
Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.


給你兩個 非空白的非負數(可能會有0)的linked lists，以降暮排序
每個節點都有一個0~9的個位數字
將每個節點數字分別加總後，以linked lists返回
返回的各節點值，僅限個位數字，進位數字加總到下一節點(需特別處理進位問題) 




 ◎ Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
*** 第二個節點相加 (4+6=10)，個位數為 0， 十位數字 加總到下一節點
所以，節點三 = 3+4+1(第二節點進位) = 8

 ◎ Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

 ◎ Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

 ◎ Constraints:
The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.




 ◎ 想法：
每個節點的值 最大=9
9+9=18，最大進位數字 =1
我們僅需要紀錄，前一節點加總後 是否有進位；有的話 這節點需 +1 
再來須注意 輸入的兩個節點長度，可能是不相同的
然後，各節點取個位數字返回即可


 ◎ C++程式碼：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
////#define DEBUG 1
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int sum = 0;
        ListNode *l3 = new ListNode;
        ListNode *node = l3;
        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;
            }
            node->next = new ListNode(sum % 10);
            node = node->next;
            sum /= 10;
        }
        return l3->next;
    }
};