【LeetCode】Add Two Numbers

题目描述：

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

直接将数字加到l1上，用一个变量来储存是否进位。若l1比l2短则将l2接到l1后面，最后直接返回l1。

代码如下：

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {if (!l1)return l2;if (!l2)return l1;int add(0), num(0);ListNode *res = l1;while (l1&&l2){num += l1->val + l2->val;if (num >= 10){l1->val = num - 10;num = 1;}else{l1->val = num;num = 0;}if (!l1->next){l1->next = l2->next;break;}if (!l2->next)break;l1 = l1->next;l2 = l2->next;}while (num){if (!l1->next)l1->next = new ListNode(0);l1 = l1->next;num += l1->val;if (num >= 10){l1->val = num - 10;num = 1;}else{l1->val = num;num = 0;}}return res;}};