第一想法是順著題目的原因，將兩鏈表分別轉化為一個數字，再將數字相加，然後把結果轉化為字符串，存到答案鏈表中。但是數據太大會溢出！

所以，要在計算一對數字的過程當中直接存儲一個結果，註意結果大於9時進位，刪去最終鏈表的最後一個節點。

/**
 * 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) {
        int carry=0;
        int x,y,z;
        ListNode* head = new ListNode(-1);
        ListNode* use = new ListNode(-1);
        ListNode* t = new ListNode(-1);
        head->next = t;
        while(l1 || l2)
        {
            if(!l1)
            {
                x 
 
= 0;
            }else x = l1->val;

            if(!l2)
            {
                y = 0;
            }else y = l2->val;

            z = x + y + carry;
            carry = 0;
            if(z>9)
            {
                carry = 1;
                z = z-10;
            }
            t->val = z;
            t
 
->next = new ListNode(-1);
            use = t;
            t = t->next;
            t->next = NULL;

            if(l1)
            {
                l1 = l1->next;
            }
            if(l2)
            {
                l2 = l2->next;
            }
        }
        if(carry)
        {
            t->val = carry;
            t->next = new ListNode(-1);
            use = t;
            t = t->next;
            t->next = NULL;
        }
        t = head->next;
        use->next = NULL;
        return t;
    }
};

2.兩數相加(Add Two Numbers)