19.1.11 LeetCode 2 Add Two Numbers

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

 1 class Solution {
 2 public:
 3     ListNode * addTwoNumbers(ListNode* l1, ListNode* l2) {
 4         int num1, num2;
 5         ListNode*ans,*oldans=NULL,*head;
 6         int inc = 0;
 7         while (l1!= NULL && l2!= NULL) {
 8             num1 = l1->val, num2 = l2->val;
 9             int plus = num1 + num2+inc;
10             inc = plus / 10;
11             ans = new ListNode(plus % 10);
12             if (oldans)
13                 oldans->next = ans;
14             else
15                 head = ans;
16             oldans = ans;
17             ans = ans->next;
18             l1 = l1->next, l2 = l2->next;
19         }
20         while (l1 != NULL) {
21             num1 = l1->val;
22             int plus = num1 + inc;
23             inc = plus / 10;
24             ans = new ListNode(plus % 10);
25             if (oldans)
26                 oldans->next = ans;
27             else
28                 head = ans;
29             oldans = ans;
30             ans = ans->next;
31             l1 = l1->next;
32         }
33         while (l2!= NULL) {
34             num1 = l2->val;
35             int plus = num1 + inc;
36             inc = plus / 10;
37             ans = new ListNode(plus % 10);
38             if (oldans)
39                 oldans->next = ans;
40             else
41                 head = ans;
42             oldans = ans;
43             ans = ans->next;
44             l2 = l2->next;
45         }
46         if (inc > 0) {
47             ans = new ListNode(inc);
48             if (oldans)
49                 oldans->next = ans;
50             else
51                 head = ans;
52         }
53         return head;
54     }
55 };

View Code

普通地做了下模拟，是不是有点太麻烦了呢？

注定失败的战争，也要拼尽全力去打赢它；就算输，也要输得足够漂亮。