[leetCode] Compare Version Numbers

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) {        if (version1 == "" || version2 == "") return 0;        String[] v1 = version1.split("\\.");        String[] v2 = version2.split("\\.");        if (v1.length == 0 || v2.length == 0) return 0;        for (int i =0; i < v1.length && i < v2.length; i++) {            int tmp1 = Integer.parseInt(v1[i]);            int tmp2 = Integer.parseInt(v2[i]);            if (tmp1 > tmp2) return 1;            else if (tmp1 < tmp2) return -1;        }        if (v1.length > v2.length) {            for (int i = v2.length; i < v1.length; i++) {                if (Integer.parseInt(v1[i]) != 0) return 1;            }        }        else if (v1.length < v2.length) {            for (int i = v1.length; i < v2.length; i++) {                if (Integer.parseInt(v2[i]) != 0) return -1;            }        }        return 0;    }}