leetcode165. Compare Version Numbers

题目要求

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

也就是说，比较版本号。

思路一：利用java API

通过split方法将版本通过.分隔开，然后将每一段版本从string转化为int进行比较

    public int compareVersion(String version1, String version2) {
        String[] v1Detail = version1.split("\\.");
        String[] v2Detail = version2.split("\\.");
        int i = 0;
        while(i<v1Detail.length && i<v2Detail.length){
            int tempV1 = Integer.parseInt(v1Detail[i]);
            int tempV2 = Integer.parseInt(v2Detail[i]);
            if(tempV1<tempV2) return -1;
            if(tempV1>tempV2) return 1;
            i++;
        }
        while(i<v2Detail.length && Integer.parseInt(v2Detail[i])==0) i++;
        while(i<v1Detail.length && Integer.parseInt(v1Detail[i])==0) i++;
        if(i>=v1Detail.length && i>=v2Detail.length) return 0;
        else if(i>=v1Detail.length) return -1;
        return 1;
    }

思路二：自己实现string转化为integer

自己实现将string转化为integer，可以通过循环的方式。这是一个基本的算法。

    public int compareVersion2(String version1, String version2){
        int i = 0;
        int j = 0;
        int length1 = version1.length();
        int length2 = version2.length();
        while(i<length1 || j<length2){
            int num1 = 0, num2 = 0;
            while(i<length1 && version1.charAt(i)!='.') num1 = num1*10+(version1.charAt(i++)-'0');
            while(j<length2 && version2.charAt(j)!='.') num2 = num2*10+(version2.charAt(j++)-'0');
            if(num1>num2) return 1;
            else if(num1<num2) return -1;
            i++;
            j++;
        }
        return 0;
    }

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