使用Java 9自带的Version类

import java.util.*;
import java.lang.module.ModuleDescriptor.Version;
class Main {
  public static void main(String[] args) {
    var versions = Arrays.asList(
      "1.0.2",
      "1.0.0-beta.2",
      "1.0.0",
      "1.0.0-beta",
      "1.0.0-alpha.12",
      "1.0.0-beta.11",
      "1.0.1",
      "1.0.11",
      "1.0.0-rc.1",
      "1.0.0-alpha.1",
      "1.1.0",
      "1.0.0-alpha.beta",
      "1.11.0",
      "1.0.0-alpha.12.ab-c",
      "0.0.1",
      "1.2.1",
      "1.0.0-alpha",
      "1.0.0.1",  // Also works with a number of sections different than 3
      "1.0.0.2",
      "2",
      "10",
      "1.0.0.10"
    );
    versions.stream()
      .map(Version::parse)
      .sorted()
      .forEach(System.out::println);
  }
}

在网上试试!

输出:

0.0.1
1.0.0-alpha
1.0.0-alpha.1
1.0.0-alpha.12
1.0.0-alpha.12.ab-c
1.0.0-alpha.beta
1.0.0-beta
1.0.0-beta.2
1.0.0-beta.11
1.0.0-rc.1
1.0.0
1.0.0.1
1.0.0.2
1.0.0.10
1.0.1
1.0.2
1.0.11
1.1.0
1.2.1
1.11.0
2
10

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。

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

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!

这篇旧文章的另一个解决方案(对那些可能有帮助的人来说):

public class Version implements Comparable<Version> {

    private String version;

    public final String get() {
        return this.version;
    }

    public Version(String version) {
        if(version == null)
            throw new IllegalArgumentException("Version can not be null");
        if(!version.matches("[0-9]+(\\.[0-9]+)*"))
            throw new IllegalArgumentException("Invalid version format");
        this.version = version;
    }

    @Override public int compareTo(Version that) {
        if(that == null)
            return 1;
        String[] thisParts = this.get().split("\\.");
        String[] thatParts = that.get().split("\\.");
        int length = Math.max(thisParts.length, thatParts.length);
        for(int i = 0; i < length; i++) {
            int thisPart = i < thisParts.length ?
                Integer.parseInt(thisParts[i]) : 0;
            int thatPart = i < thatParts.length ?
                Integer.parseInt(thatParts[i]) : 0;
            if(thisPart < thatPart)
                return -1;
            if(thisPart > thatPart)
                return 1;
        }
        return 0;
    }

    @Override public boolean equals(Object that) {
        if(this == that)
            return true;
        if(that == null)
            return false;
        if(this.getClass() != that.getClass())
            return false;
        return this.compareTo((Version) that) == 0;
    }

}

Version a = new Version("1.1");
Version b = new Version("1.1.1");
a.compareTo(b) // return -1 (a<b)
a.equals(b)    // return false

Version a = new Version("2.0");
Version b = new Version("1.9.9");
a.compareTo(b) // return 1 (a>b)
a.equals(b)    // return false

Version a = new Version("1.0");
Version b = new Version("1");
a.compareTo(b) // return 0 (a=b)
a.equals(b)    // return true

Version a = new Version("1");
Version b = null;
a.compareTo(b) // return 1 (a>b)
a.equals(b)    // return false

List<Version> versions = new ArrayList<Version>();
versions.add(new Version("2"));
versions.add(new Version("1.0.5"));
versions.add(new Version("1.01.0"));
versions.add(new Version("1.00.1"));
Collections.min(versions).get() // return min version
Collections.max(versions).get() // return max version

// WARNING
Version a = new Version("2.06");
Version b = new Version("2.060");
a.equals(b)    // return false

编辑:

@daiscog:谢谢你的评论，这段代码是为Android平台开发的，由谷歌推荐，方法“匹配”检查整个字符串，不像Java使用监管模式。(Android文档- JAVA文档)

下面是一个优化的实现:

public static final Comparator<CharSequence> VERSION_ORDER = new Comparator<CharSequence>() {

  @Override
  public int compare (CharSequence lhs, CharSequence rhs) {
    int ll = lhs.length(), rl = rhs.length(), lv = 0, rv = 0, li = 0, ri = 0;
    char c;
    do {
      lv = rv = 0;
      while (--ll >= 0) {
        c = lhs.charAt(li++);
        if (c < '0' || c > '9')
          break;
        lv = lv*10 + c - '0';
      }
      while (--rl >= 0) {
        c = rhs.charAt(ri++);
        if (c < '0' || c > '9')
          break;
        rv = rv*10 + c - '0';
      }
    } while (lv == rv && (ll >= 0 || rl >= 0));
    return lv - rv;
  }

};

结果:

"0.1" - "1.0" = -1
"1.0" - "1.0" = 0
"1.0" - "1.0.0" = 0
"10" - "1.0" = 9
"3.7.6" - "3.7.11" = -5
"foobar" - "1.0" = -1