LeetCode OJ

题目：

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

解题思路：

直接一个个的加就行了，注意进位和边界处理。

代码：

 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         if (l1 == NULL) return l2;
13         if (l2 == NULL) return l1;
14         
15         ListNode* head = new ListNode((l1->val + l2->val) % 10);
16         int carry = (l1->val + l2->val) / 10;
17         ListNode* p1 = l1->next, *p2 = l2->next;
18         
19         ListNode* cur = head;
20         while (p1 != NULL && p2 != NULL) {
21             int tmp = p1->val + p2->val + carry;
22             cur->next = new ListNode(tmp % 10);
23             carry = tmp / 10;
24             p1 = p1->next;
25             p2 = p2->next;
26             cur = cur->next;
27         }
28         while (p1 != NULL) {
29             int tmp = p1->val + 0 + carry;
30             cur->next = new ListNode(tmp % 10);
31             carry = tmp / 10;
32             p1 = p1->next;
33             cur = cur->next;
34         }
35         while (p2 != NULL) {
36             int tmp = p2->val + 0 + carry;
37             cur->next = new ListNode(tmp % 10);
38             carry = tmp / 10;
39             p2 = p2->next;
40             cur = cur->next;
41         }
42         if (carry) {
43             cur->next = new ListNode(carry);
44         }
45         return head;
46     }
47 };