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

大非负整数相加，注意数字是怎么表示的就好，链表头是低位，尾是高位，另外最后一位记得处理，也不要忘了0的情况。

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {12         ListNode* p1 = l1, *p2 = l2;13         ListNode* ret = NULL;14         ListNode* p = ret;15         // 逐位相加16         while (p1 != NULL && p2 != NULL) {17             ListNode* tmp = new ListNode(p1->val + p2->val);18             // 判断是不是链表头19             if (ret == NULL) {20                 ret = tmp;21                 p = ret;22             } else {23                 p->next = tmp;24                 p = tmp;25             }26             p1 = p1->next;27             p2 = p2->next;28         }29         30         if (p1 != NULL) {31             p->next = p1;32         }33         if (p2 != NULL) {34             p->next = p2;35         }36         37         // 为空时要返回038         if (ret == NULL)39             return new ListNode(0);40         41         // 处理进位问题42         p = ret;43         while (p->next != NULL) {44             p->next->val += p->val / 10;45             p->val = p->val % 10;46             p = p->next;47         }48         // 处理最高位49         if (p->val >= 10) {50             ListNode* tmp = new ListNode(p->val / 10);51             p->next = tmp;52             p->val = p->val % 10;53         }54         55         return ret;56         57     }58 };