Add two numbers [LeetCode]

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

Summary: Be careful about the last carry.

 1     ListNode * handler(int sum, int * carry, ListNode * result, ListNode * * new_head){
 2         if(result == NULL){
 3             result = new ListNode(sum % 10);
 4             (*new_head) = result;
 5         }else{
 6             result->next = new ListNode(sum % 10);
 7             result = result->next;
 8         }
 9             
10         (*carry) = sum / 10; 
11         return result;
12     }
13     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
14         ListNode * new_head = NULL;
15         ListNode * result = NULL;
16         int carry = 0;
17         while(l1 != NULL || l2 != NULL){
18             if(l1 == NULL){
19                 int sum = l2->val + carry;
20                 result = handler(sum, &carry, result, &new_head);
21                 l2 = l2->next;
22                 continue;
23             }
24             
25             if(l2 == NULL){
26                 int sum = l1->val + carry;
27                 result = handler(sum, &carry, result, &new_head);
28                 l1 = l1->next;
29                 continue;
30             }
31             
32             int sum = l1->val + l2->val + carry;
33             result = handler(sum, &carry, result, &new_head);
34             l1 = l1->next;
35             l2 = l2->next;
36         }
37         
38         if(carry != 0 )
39             result->next = new ListNode(carry);
40             
41         return new_head;
42     }

 

posted @ 2013-11-25 15:51  假日笛声  阅读(173)  评论(0)  编辑  收藏  举报