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

[Solution]

本题比较简单，只是简单的链表操作，只是不要忘记最后进位情况。

1 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)  2     { 3         ListNode dummy, *p = NULL; 4         int carry = 0; 5          6         p = &dummy; 7         while (l1 != NULL | l2 != NULL) 8         { 9             int sum = (l1 == NULL) ? 0 : l1->val + (l2 == NULL) ? 0 : l2->val + carry;10             p->next = new ListNode(sum % 10);11             p = p->next;12             carry = sum / 10;13             if (l1)14                 l1 = l1->next;15             if (l2)16                 l2 = l2->next;17         }18         if (carry)19             p->next = new ListNode(carry);20         21         return dummy.next;22     }