第一想法是顺着题目的原因，将两链表分别转化为一个数字，再将数字相加，然后把结果转化为字符串，存到答案链表中。但是数据太大会溢出！

所以，要在计算一对数字的过程当中直接存储一个结果，注意结果大于9时进位，删去最终链表的最后一个节点。

/** * 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;    }};