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Leetcode 2: Add Two Numbers

时间:2017-11-04 13:20:12      阅读:100      评论:0      收藏:0      [点我收藏+]

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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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

 


 
 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public int val;
 5  *     public ListNode next;
 6  *     public ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
11         ListNode ret = null;
12         ListNode curNode = null;
13         int advance = 0;
14         
15         while (l1 != null || l2 != null)
16         {
17             int cur = advance;
18             
19             if (l1 != null)
20             {
21                 cur += l1.val;
22                 l1 = l1.next;
23             }
24                         
25             if (l2 != null)
26             {
27                 cur += l2.val;
28                 l2 = l2.next;
29             }
30             
31             if (ret == null)
32             {
33                 ret = new ListNode(cur % 10);
34                 curNode = ret;
35             }
36             else 
37             {
38                 curNode.next = new ListNode(cur % 10);
39                 curNode = curNode.next;
40             }
41             
42             advance = cur / 10;
43         }
44         
45         if (advance != 0) curNode.next = new ListNode(advance);
46         
47         return ret;
48     }
49 }

 

 

Leetcode 2: Add Two Numbers

标签:ext   put   exce   cep   red   rip   ott   blog   turn   

原文地址:http://www.cnblogs.com/liangmou/p/7782953.html

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