【Leetcode】Compare Version Numbers

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题目链接：https://leetcode.com/problems/compare-version-numbers/

题目：

Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

算法：

[java] view plain copy

 

1. public int compareVersion(String version1, String version2) {  
2.     String[] v1 = version1.split("\\.");  
3.     String[] v2 = version2.split("\\.");  
4.     int length = Math.max(v1.length, v2.length);  
5.     int vv2[] = new int[length];  
6.     int vv1[] = new int[length];  
7.     for (int i = 0; i < v1.length; i++) {  
8.         vv1[i] = Integer.parseInt(v1[i]);  
9.     }  
10.     for (int i = 0; i < v2.length; i++) {  
11.         vv2[i] = Integer.parseInt(v2[i]);  
12.     }  
13.       
14.     int flag = -1;  
15.     int i = 0, j = 0;  
16.     while (i < length) {  
17.         if (vv1[i] > vv2[i]) {  
18.             flag = 1;  
19.             break;  
20.         } else if (vv1[i] < vv2[j]) {  
21.             flag = -1;  
22.             break;  
23.         } else {  
24.             flag = 0;  
25.             i++;  
26.             j++;  
27.         }  
28.     }  
29.     return flag;  
30. }  

【Leetcode】Compare Version Numbers

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原文地址：http://blog.csdn.net/yeqiuzs/article/details/51590916