165. Compare Version Numbers

最新推荐文章于 2022-02-25 21:50:14 发布

mingyangdai

最新推荐文章于 2022-02-25 21:50:14 发布

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分类专栏：数据结构&算法文章标签：数据结构算法 data structure algorithm

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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

比较版本号

做出来的和答案基本一致只是修改了几次

public int compareVersion(String version1, String version2) {
    String[] levels1 = version1.split("\\.");
    String[] levels2 = version2.split("\\.");
    
    int length = Math.max(levels1.length, levels2.length);
    for (int i=0; i<length; i++) {
        Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0;
        Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0;
        int compare = v1.compareTo(v2);
        if (compare != 0) {
                return compare;
        }
    }
    
    return 0;
}

注意版本号分割后长度不同时的处理方式