题目：
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 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
解题思路：
- 将数字 342 反序2-4-3，然后使用链表保存 2-4-3
- 将数字 465 反序5-6-4，然后使用链表保存 5-6-7
- 将 342 + 465 的和 807 反序，然后使用链表保存 7-0-8
- 这道题就是加法题，需要注意的是数字相加产生的进位处理
- 函数的返回值应该是链表的头部
解题代码：
/**
* 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 *ret_list = new ListNode(-1);
ListNode *list_head_node = ret_list;
int carry = 0;
while (l1 || l2) {
int data1 = l1 ? l1->val : 0;
int data2 = l2 ? l2->val : 0;
int sum = data1 + data2 + carry;
carry = sum / 10;
ret_list->next = new ListNode(sum % 10);
ret_list = ret_list->next;
l1 = l1 ? l1->next : NULL;
l2 = l2 ? l2->next : NULL;
}
if (carry != 0) {
ret_list->next = new ListNode(carry);
}
return list_head_node->next;
}
};
