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

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!

其他回答

对于我的项目,我使用我的公共版本库https://github.com/raydac/commons-version 它包含两个辅助类-用于解析版本(解析后的版本可以与另一个版本对象进行比较,因为它是可比的)和VersionValidator,它允许检查一些表达式的版本,如!=ide-1.1.1,>idea-1.3.4-SNAPSHOT;<1.2.3

使用Maven真的很简单:

import org.apache.maven.artifact.versioning.DefaultArtifactVersion;

DefaultArtifactVersion minVersion = new DefaultArtifactVersion("1.0.1");
DefaultArtifactVersion maxVersion = new DefaultArtifactVersion("1.10");

DefaultArtifactVersion version = new DefaultArtifactVersion("1.11");

if (version.compareTo(minVersion) < 0 || version.compareTo(maxVersion) > 0) {
    System.out.println("Sorry, your version is unsupported");
}

您可以从这个页面获得Maven Artifact的正确依赖项字符串:

<dependency>
<groupId>org.apache.maven</groupId>
<artifactId>maven-artifact</artifactId>
<version>3.0.3</version>
</dependency>

如果你的项目中已经有Jackson,你可以使用com.fasterxml.jackson.core.Version:

import com.fasterxml.jackson.core.Version;
import org.junit.Test;

import static org.junit.Assert.assertTrue;

public class VersionTest {

    @Test
    public void shouldCompareVersion() {
        Version version1 = new Version(1, 11, 1, null, null, null);
        Version version2 = new Version(1, 12, 1, null, null, null);
        assertTrue(version1.compareTo(version2) < 0);
    }
}

用点作为分隔符对字符串进行标记,然后从左边开始并排比较整数转换。

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