Add Two Numbers

Description:

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

Definition for singly-linked list:

struct ListNode {
      int val;
      ListNode *next;
      ListNode(int x) : val(x), next(NULL) {}
 };

Code:

 1  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
 2         size_t lengthL1 = 0;
 3         size_t lengthL2 = 0;
 4         ListNode *result = NULL;
 5         ListNode *pL1 = l1;
 6         ListNode *pL2 = l2;
 7         while (pL1)
 8         {
 9             lengthL1++;
10             pL1 = pL1->next;
11         }
12         while (pL2)
13         {
14             lengthL2++;
15             pL2 = pL2->next;
16         }
17         
18         ListNode *longList = (lengthL1 >= lengthL2)?l1:l2;
19         ListNode *shortList = (lengthL1 < lengthL2)?l1:l2;
20         
21         //设置进位标志
22         int flag = 0;
23         int x;
24         ListNode *p = NULL;
25         while (shortList)
26         {
27             x = flag + longList->val + shortList->val;
28             flag = x/10;
29             if (result==NULL)
30             {
31                 result = new ListNode ( x%10 );
32                 p = result;
33             }
34             else
35             {
36                 p->next = new ListNode (x%10);
37                 p = p->next;
38             }
39             shortList = shortList->next;
40             longList = longList->next;
41         }
42          while (longList)
43         {
44             if (flag == 0)
45             {
46                    p->next = longList; 
47                    return result;
48             }
49             else
50             {
51                 x = flag + longList->val;
52                 flag = x/10;
53                 p->next = new ListNode (x%10);
54                 p = p->next;
55                 longList = longList->next;
56              }
57         }
58             if (flag==1)
59             {
60                 p->next = new ListNode (1);
61             }
62             return result;
63     }

 

posted @ 2015-06-14 13:15  Rosanne  阅读(175)  评论(0)  编辑  收藏  举报