[leetcode] Compare Version Numbers

问题描述：

Compare two version numbers version1 andversion1.
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

基本思想：

从高位到低位递归解决。 注意特殊case （1,1.0）;

代码：

    public int compareVersion(String version1, String version2) { //java        if(version1.equals(version2))            return 0;                    int fversion1 , fversion2;        String sversion1,sversion2;        if(version1.contains(".")){            int pos = version1.indexOf(".");            fversion1 = Integer.valueOf(version1.substring(0,pos));            sversion1 = version1.substring(pos+1,version1.length());        }        else {            fversion1 = Integer.valueOf(version1);            sversion1 = "0";        }                if(version2.contains(".")){            int pos = version2.indexOf(".");            fversion2 = Integer.valueOf(version2.substring(0,pos));            sversion2 = version2.substring(pos+1,version2.length());        }        else {            fversion2 = Integer.valueOf(version2);            sversion2 = "0";        }                if(fversion1 > fversion2)            return 1;        else if(fversion1 < fversion2)            return -1;        else return compareVersion(sversion1, sversion2);    }