[Java]Leetcode165 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

题意是：比较版本号的大小，假设版本号只含有数字和小数点。这一题首先可能想到的是字符转换为整型比较大小，但是版本号中可以有多个小数点，这一点就不符合了。于是想到用split()方法，以小数点为分界符，分成一个字符数组，再一一比较。

代码如下：

 public int compareVersion(String version1, String version2) {        String[] ver1=version1.split("\\.");        String[] ver2=version2.split("\\.");        int len1=ver1.length;        int len2=ver2.length;        int len=0;        if(len2<=len1)len=len2;        else len=len1;        int num1=0;        int num2=0;        int i=0;        while(i<len)        {           num1=Integer.valueOf(ver1[i]);           num2=Integer.valueOf(ver2[i]);           if(num1==num2)i++;           else if(num1<num2)return -1;           else return 1;        }        while(len1>i)//有小数点，但是小数点后的值为0的情况，不为0返回1，都为0就return 0.        {          if(Integer.valueOf(ver1[i++])!=0)return 1;        }        i=len;        while(len2>i)        {           if(Integer.valueOf(ver2[i++])!=0)return -1;          }        return 0;    }