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：链表的基本操作要会，基本就是实现十进制加法。
/** * 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 l(0), *p = &l; int sum = 0; int reminder = 0; while(l1 || l2 || reminder){ sum = (l1?l1->val:0) + (l2?l2->val:0) + reminder; reminder = sum/10; sum = sum ; p->next = new ListNode(sum); p = p->next; l1 = l1?l1->next:l1; l2 = l2?l2->next:l2; } return l.next; } };