[LeetCode] Compare Version Numbers

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：