2. Add Two Numbers

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：

1. 将数字 342 反序2-4-3，然后使用链表保存 2-4-3
2. 将数字 465 反序5-6-4，然后使用链表保存 5-6-7
3. 将 342 + 465 的和 807 反序，然后使用链表保存 7-0-8
4. 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
5. 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;
        
    }
};

参考

1. https://leetcode.com/problems/add-two-numbers/description/