Compare Version Numbers

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

Solution: 

先根据点split strings，存在String［］里面。

从高位（第一个）开始对比，如果version1大，则返回1，如果version2大，则返回-1。

如果version1比较长，查看长出来的部分是否大于0， 是则返回1.

如果version2比较长，查看长出部分时候大于0， 是则返回-1。

其他情况返回0.

复杂度是O(n).

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");
        
        for(int i = 0; i < v1.length || i < v2.length; i++) {
            if(i < v1.length && i < v2.length) {
                if(Integer.valueOf(v1[i]) > Integer.valueOf(v2[i])) {
                    return 1;
                }
                if(Integer.valueOf(v1[i]) < Integer.valueOf(v2[i])) {
                    return -1;
                }
            }else if(i < v1.length) {  //version1 is longer than version2
                if(Integer.valueOf(v1[i]) > 0) {
                    return 1;
                }  
            }else {  //version2 is longer than version1
                if(Integer.valueOf(v2[i]) > 0) {
                    return -1;
                }
            }
        }
        return 0;
    }
}