Leetcode OJ: Add Two Numbers

Add Two Numbers

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

大非负整数相加，注意数字是怎么表示的就好，链表头是低位，尾是高位，另外最后一位记得处理，也不要忘了0的情况。

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
12         ListNode* p1 = l1, *p2 = l2;
13         ListNode* ret = NULL;
14         ListNode* p = ret;
15         // 逐位相加
16         while (p1 != NULL && p2 != NULL) {
17             ListNode* tmp = new ListNode(p1->val + p2->val);
18             // 判断是不是链表头
19             if (ret == NULL) {
20                 ret = tmp;
21                 p = ret;
22             } else {
23                 p->next = tmp;
24                 p = tmp;
25             }
26             p1 = p1->next;
27             p2 = p2->next;
28         }
29         
30         if (p1 != NULL) {
31             p->next = p1;
32         }
33         if (p2 != NULL) {
34             p->next = p2;
35         }
36         
37         // 为空时要返回0
38         if (ret == NULL)
39             return new ListNode(0);
40         
41         // 处理进位问题
42         p = ret;
43         while (p->next != NULL) {
44             p->next->val += p->val / 10;
45             p->val = p->val % 10;
46             p = p->next;
47         }
48         // 处理最高位
49         if (p->val >= 10) {
50             ListNode* tmp = new ListNode(p->val / 10);
51             p->next = tmp;
52             p->val = p->val % 10;
53         }
54         
55         return ret;
56         
57     }
58 };