[leet code 165]Compare Version Numbers

1 题目

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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

2 思路

好吧，这题很蛋疼，各种情况。有1.0与1比较的，有1.2.3与1.2比较的，考虑到各种情况，很复杂。还发现java里面split函数输入是正则表达式，要用"."符号分割的话，得这样.split"\."，才能分割。

网上查了查，有一个思路是用递归解决的，http://www.tuicool.com/articles/3QV7NvV。完美解决 比较 1.0与1的问题。

我感觉实际操作中，会规定死版本的格式吧。如1.2.3与1.0.0。另外，我看实际中的一个代码，只要发现两个版本的字符串不一样，就提醒用户更新版本－ －。

3 代码

    public int compareVersion(String version1, String version2) {    
        if(version1.equals(version2))
            return 0;
        int fversion1 , fversion2;//最左边字符串代表的数字
        String sversion1,sversion2;//剔除.号左边的数字剩下的字符串
        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";//预防比较1.0与1这种情况
        }
        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);
    }