題目：
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
解題思路：
- 將數(shù)字 342 反序2-4-3束昵，然后使用鏈表保存 2-4-3
- 將數(shù)字 465 反序5-6-4，然后使用鏈表保存 5-6-7
- 將 342 + 465 的和 807 反序醒颖，然后使用鏈表保存 7-0-8
- 這道題就是加法題妻怎，需要注意的是數(shù)字相加產(chǎn)生的進(jìn)位處理
- 函數(shù)的返回值應(yīng)該是鏈表的頭部
解題代碼：
/**
* 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;
}
};
