leetcode165.Compare Version Numbers

发表于 更新于 分类于 ，

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
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。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
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;
}