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.

思路：善於使用條件運算子，注意進位，此處新建了一個連結串列用來儲存計算結果。我個人覺得可以不用新建一個連結串列，直接覆蓋計算結果，

修改next指標的指向應該也可以。

/**
 * 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 preHead(0), *p = &preHead;
        int carry_bit = 0;
        while (l1 || l2 || carry_bit) 
        {
            int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry_bit;
            carry_bit = sum / 10;
            p->next = new ListNode(sum % 10);
            p = p->next;
            l1 = l1 ? l1->next : l1;
            l2 = l2 ? l2->next : l2;
        }
        return preHead.next;
    }
};