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Question：
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
1、题型分类：
2、思路：
3、时间复杂度：
4、代码：
public class Solution { public int compareVersion(String version1, String version2) { String[] v1=version1.split("\\."); String[] v2=version2.split("\\."); int x=Math.max(v1.length, v2.length); int [] i1=convertStringArrayToIntegerArray(v1,x); int [] i2=convertStringArrayToIntegerArray(v2,x); for(int i=0;i<x;i++) { if(i1[i]>i2[i]) return 1; else if(i1[i]<i2[i]) return -1; } return 0; } public int[] convertStringArrayToIntegerArray(String[] str,int l) { int len=str.length; int [] r=new int[l]; for(int i=0;i<len;i++) { r[i]=Integer.parseInt(str[i]); } return r; } }
5、优化：
6、扩展：
[LeetCode] Compare Version Numbers
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原文地址：http://www.cnblogs.com/maydow/p/4644086.html
