Compare two version numbers version1 and version1.
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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");
        for(String s:v1){
        	System.out.println(s);
        }
        for(int i=0;i<v1.length&&i<v2.length;i++){
        	int n1 = Integer.parseInt(v1[i]),n2 = Integer.parseInt(v2[i]);
        	if(n1>n2){
        		return 1;
        	}else if(n1<n2){
        		return -1;
        	}
        }
        if(v1.length<v2.length){
        	for(int i=v1.length;i<v2.length;i++){
        		if(Integer.parseInt(v2[i])>0) return -1;
        	}
        }else if(v1.length>v2.length){
        	for(int i=v2.length;i<v1.length;i++){
        		if(Integer.parseInt(v1[i])>0) return 1;
        	}
        }
        return 0;
    }
}

[LeetCode]Compare Version Numbers