[Leetcode] Compare Version Numbers

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

Assume that the version strings is non-empty and contain only digits and the.Character.
The.Character does not represent a decimal point and was used to separate number sequences.
For instance,2.5is not "both and a half" or "half-to-version three", it is the fifth Second-level revision of the second first-level re Vision.

Here are 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