LeetCode OJ：Add Two Numbers (相加链表之数)

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

将两个链表上的数相加就可以，大于10进一位，注意下相加时候的细节就可以了，我这里

吧prev节点记录下来，这样最后多生成节点的时候便于将最后一个多余的节点删除：

 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 * curr = new ListNode(0);
13         ListNode * root = curr;
14         ListNode * prev = curr;
15         int currVal = 0;
16         while(l1 != NULL && l2 != NULL){
17             currVal = l1->val + l2->val;
18             curr->val += currVal;
19             curr->next = new ListNode(curr->val/10);
20             curr->val%=10;
21             prev = curr;
22             curr = curr->next;
23             l1 = l1->next;
24             l2 = l2->next;
25         }
26         while(l1 != NULL){
27             curr->val += l1->val;
28             curr->next = new ListNode(curr->val/10);
29             curr->val %= 10;
30             prev = curr;
31             curr = curr->next;
32             l1 = l1->next;
33         }
34         while(l2 != NULL){
35             curr->val += l2->val;
36             curr->next = new ListNode(curr->val/10);
37             curr->val %= 10;
38             prev = curr;
39             curr = curr->next;
40             l2 = l2->next;
41         }
42         if(curr->val == 0){
43             prev->next = NULL;
44             delete curr;
45         }
46         return root;
47     }
48 };

感觉写的有点麻烦，应该有很多的重复代码可以改正，但是我暂时找不出来了，先这样吧。

更新下，以前脑子抽了写出了那样的代码。其实三个while循环都可以放到一个while中，下面用java在写一起，方法还是类似的：

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
11         ListNode tmp = new ListNode(0);
12         ListNode head = tmp;
13         int carry = 0;
14         while(l1!=null || l2!=null || carry != 0){
15             int val = ((l1 != null)?l1.val:0) + ((l2!=null)?l2.val:0) + carry;
16             carry = carry/10 + val/10;
17             val %= 10;
18             tmp.next = new ListNode(val);
19             tmp = tmp.next;
20             l1 = l1!=null ? l1.next : l1;
21             l2 = l2!=null ? l2.next : l2;
22         }
23         return head.next;
24     }
25 }