科特林：

@kotlin.jvm.Throws(InvalidParameterException::class)
fun String.versionCompare(remoteVersion: String?): Int {
    val remote = remoteVersion?.splitToSequence(".")?.toList() ?: return 1
    val local = this.splitToSequence(".").toList()

    if(local.filter { it.toIntOrNull() != null }.size != local.size) throw InvalidParameterException("version invalid: $this")
    if(remote.filter { it.toIntOrNull() != null }.size != remote.size) throw InvalidParameterException("version invalid: $remoteVersion")

    val totalRange = 0 until kotlin.math.max(local.size, remote.size)
    for (i in totalRange) {
        if (i < remote.size && i < local.size) {
            val result = local[i].compareTo(remote[i])
            if (result != 0) return result
        } else (
            return local.size.compareTo(remote.size)
        )
    }
    return 0
}

由于本页上没有答案能很好地处理混合文本，我做了自己的版本:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

class Main {
    static double parseVersion(String v) {
        if (v.isEmpty()) {
            return 0;
        }
        Pattern p = Pattern.compile("^(\\D*)(\\d*)(\\D*)$");
        Matcher m = p.matcher(v);
        m.find();
        if (m.group(2).isEmpty()) {
            // v1.0.0.[preview]
            return -1;
        }
        double i = Integer.parseInt(m.group(2));
        if (!m.group(3).isEmpty()) {
            // v1.0.[0b]
            i -= 0.1;
        }
        return i;
    }

    public static int versionCompare(String str1, String str2) {
        String[] v1 = str1.split("\\.");
        String[] v2 = str2.split("\\.");
        int i = 0;
        for (; i < v1.length && i < v2.length; i++) {
            double iv1 = parseVersion(v1[i]);
            double iv2 = parseVersion(v2[i]);

            if (iv1 != iv2) {
                return iv1 - iv2 < 0 ? -1 : 1;
            }
        }
        if (i < v1.length) {
            // "1.0.1", "1.0"
            double iv1 = parseVersion(v1[i]);
            return iv1 < 0 ? -1 : (int) Math.ceil(iv1);
        }
        if (i < v2.length) {
            double iv2 = parseVersion(v2[i]);
            return -iv2 < 0 ? -1 : (int) Math.ceil(iv2);
        }
        return 0;
    }


    public static void main(String[] args) {
        System.out.println("versionCompare(v1.0.0, 1.0.0)");
        System.out.println(versionCompare("v1.0.0", "1.0.0")); // 0

        System.out.println("versionCompare(v1.0.0b, 1.0.0)");
        System.out.println(versionCompare("v1.0.0b", "1.0.0")); // -1

        System.out.println("versionCompare(v1.0.0.preview, 1.0.0)");
        System.out.println(versionCompare("v1.0.0.preview", "1.0.0")); // -1

        System.out.println("versionCompare(v1.0, 1.0.0)");
        System.out.println(versionCompare("v1.0", "1.0.0")); // 0

        System.out.println("versionCompare(ver1.0, 1.0.1)");
        System.out.println(versionCompare("ver1.0", "1.0.1")); // -1
    }
}

不过，在需要比较“alpha”和“beta”的情况下，它仍然不够。

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 int compare(String v1, String v2) {
        v1 = v1.replaceAll("\\s", "");
        v2 = v2.replaceAll("\\s", "");
        String[] a1 = v1.split("\\.");
        String[] a2 = v2.split("\\.");
        List<String> l1 = Arrays.asList(a1);
        List<String> l2 = Arrays.asList(a2);


        int i=0;
        while(true){
            Double d1 = null;
            Double d2 = null;

            try{
                d1 = Double.parseDouble(l1.get(i));
            }catch(IndexOutOfBoundsException e){
            }

            try{
                d2 = Double.parseDouble(l2.get(i));
            }catch(IndexOutOfBoundsException e){
            }

            if (d1 != null && d2 != null) {
                if (d1.doubleValue() > d2.doubleValue()) {
                    return 1;
                } else if (d1.doubleValue() < d2.doubleValue()) {
                    return -1;
                }
            } else if (d2 == null && d1 != null) {
                if (d1.doubleValue() > 0) {
                    return 1;
                }
            } else if (d1 == null && d2 != null) {
                if (d2.doubleValue() > 0) {
                    return -1;
                }
            } else {
                break;
            }
            i++;
        }
        return 0;
    }

我创建了一个简单的实用程序，使用语义版本约定在Android平台上比较版本。所以它只适用于X.Y.Z (Major.Minor.Patch)格式的字符串，其中X、Y和Z是非负整数。你可以在我的GitHub上找到它。

方法version . compareversions (String v1, String v2)比较两个版本字符串。如果版本相等则返回0，如果版本v1在版本v2之前则返回1，如果版本v1在版本v2之后则返回-1，如果版本格式无效则返回-2。

您需要规范化版本字符串，以便对它们进行比较。类似的

import java.util.regex.Pattern;

public class Main {
    public static void main(String... args) {
        compare("1.0", "1.1");
        compare("1.0.1", "1.1");
        compare("1.9", "1.10");
        compare("1.a", "1.9");
    }

    private static void compare(String v1, String v2) {
        String s1 = normalisedVersion(v1);
        String s2 = normalisedVersion(v2);
        int cmp = s1.compareTo(s2);
        String cmpStr = cmp < 0 ? "<" : cmp > 0 ? ">" : "==";
        System.out.printf("'%s' %s '%s'%n", v1, cmpStr, v2);
    }

    public static String normalisedVersion(String version) {
        return normalisedVersion(version, ".", 4);
    }

    public static String normalisedVersion(String version, String sep, int maxWidth) {
        String[] split = Pattern.compile(sep, Pattern.LITERAL).split(version);
        StringBuilder sb = new StringBuilder();
        for (String s : split) {
            sb.append(String.format("%" + maxWidth + 's', s));
        }
        return sb.toString();
    }
}

打印

'1.0' < '1.1' '1.0.1' < '1.1' '1.9' < '1.10' “1。A ' > '1.9'