是否有比较版本号的标准习语?我不能直接使用String compareTo,因为我还不知道点释放的最大数量是多少。我需要比较版本,并有以下保持正确:

1.0 < 1.1
1.0.1 < 1.1
1.9 < 1.10

当前回答

基于https://stackoverflow.com/a/27891752/2642478

class Version(private val value: String) : Comparable<Version> {
    private val splitted by lazy { value.split("-").first().split(".").map { it.toIntOrNull() ?: 0 } }

    override fun compareTo(other: Version): Int {
        for (i in 0 until maxOf(splitted.size, other.splitted.size)) {
            val compare = splitted.getOrElse(i) { 0 }.compareTo(other.splitted.getOrElse(i) { 0 })
            if (compare != 0)
                return compare
        }
        return 0
    }
}

你可以用like:

    System.err.println(Version("1.0").compareTo( Version("1.0")))
    System.err.println(Version("1.0") < Version("1.1"))
    System.err.println(Version("1.10") > Version("1.9"))
    System.err.println(Version("1.10.1") > Version("1.10"))
    System.err.println(Version("0.0.1") < Version("1"))

其他回答

public int CompareVersions(String version1, String version2)
{
    String[] string1Vals = version1.split("\\.");
    String[] string2Vals = version2.split("\\.");

    int length = Math.max(string1Vals.length, string2Vals.length);

    for (int i = 0; i < length; i++)
    {
        Integer v1 = (i < string1Vals.length)?Integer.parseInt(string1Vals[i]):0;
        Integer v2 = (i < string2Vals.length)?Integer.parseInt(string2Vals[i]):0;

        //Making sure Version1 bigger than version2
        if (v1 > v2)
        {
            return 1;
        }
        //Making sure Version1 smaller than version2
        else if(v1 < v2)
        {
            return -1;
        }
    }

    //Both are equal
    return 0;
}

最好的方法是重用现有代码, 使用Maven的ComparableVersion类

优点:

Apache许可证,版本2.0, 测试, 在多个项目中使用(复制),如spring-security-core, jboss等 多个特性 它已经是java.lang。可比的了 只是复制粘贴一个类,没有第三方依赖

不要包含对maven-artifact的依赖项,因为那会拉动各种传递依赖项

也许有人会对我的解决方案感兴趣:

class Version private constructor(private val versionString: String) : Comparable<Version> {

    private val major: Int by lazy { versionString.split(".")[0].toInt() }

    private val minor: Int by lazy { versionString.split(".")[1].toInt() }

    private val patch: Int by lazy {
        val splitArray = versionString.split(".")

        if (splitArray.size == 3)
            splitArray[2].toInt()
        else
            0
    }

    override fun compareTo(other: Version): Int {
        return when {
            major > other.major -> 1
            major < other.major -> -1
            minor > other.minor -> 1
            minor < other.minor -> -1
            patch > other.patch -> 1
            patch < other.patch -> -1
            else -> 0
        }
    }

    override fun equals(other: Any?): Boolean {
        if (other == null || other !is Version) return false
        return compareTo(other) == 0
    }

    override fun hashCode(): Int {
        return major * minor * patch
    }

    companion object {
        private fun doesContainsVersion(string: String): Boolean {
            val versionArray = string.split(".")

            return versionArray.size in 2..3
                    && versionArray[0].toIntOrNull() != null
                    && versionArray[1].toIntOrNull() != null
                    && (versionArray.size == 2 || versionArray[2].toIntOrNull() != null)
        }

        fun from(string: String): Version? {
            return if (doesContainsVersion(string)) {
                Version(string)
            } else {
                null
            }
        }
    }
}

用法:

val version1 = Version.from("3.2")
val version2 = Version.from("3.2.1")
version1 <= version2

我喜欢@Peter Lawrey的想法,我把它扩展到更远的范围:

    /**
    * Normalize string array, 
    * Appends zeros if string from the array
    * has length smaller than the maxLen.
    **/
    private String normalize(String[] split, int maxLen){
        StringBuilder sb = new StringBuilder("");
        for(String s : split) {
            for(int i = 0; i<maxLen-s.length(); i++) sb.append('0');
            sb.append(s);
        }
        return sb.toString();
    }

    /**
    * Removes trailing zeros of the form '.00.0...00'
    * (and does not remove zeros from, say, '4.1.100')
    **/
    public String removeTrailingZeros(String s){
        int i = s.length()-1;
        int k = s.length()-1;
        while(i >= 0 && (s.charAt(i) == '.' || s.charAt(i) == '0')){
          if(s.charAt(i) == '.') k = i-1;
          i--;  
        } 
        return s.substring(0,k+1);
    }

    /**
    * Compares two versions(works for alphabets too),
    * Returns 1 if v1 > v2, returns 0 if v1 == v2,
    * and returns -1 if v1 < v2.
    **/
    public int compareVersion(String v1, String v2) {

        // Uncomment below two lines if for you, say, 4.1.0 is equal to 4.1
        // v1 = removeTrailingZeros(v1);
        // v2 = removeTrailingZeros(v2);

        String[] splitv1 = v1.split("\\.");
        String[] splitv2 = v2.split("\\.");
        int maxLen = 0;
        for(String str : splitv1) maxLen = Math.max(maxLen, str.length());
        for(String str : splitv2) maxLen = Math.max(maxLen, str.length());
        int cmp = normalize(splitv1, maxLen).compareTo(normalize(splitv2, maxLen));
        return cmp > 0 ? 1 : (cmp < 0 ? -1 : 0);
    }

希望它能帮助到别人。它通过了interviewbit和leetcode中的所有测试用例(需要取消compareVersion函数中的两行注释)。

很容易测试!

public static int compareVersions(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;
}