我自己写了一个小函数。更简单地使用列表

 public static boolean checkVersionUpdate(String olderVerison, String newVersion) {
        if (olderVerison.length() == 0 || newVersion.length() == 0) {
            return false;
        }
        List<String> newVerList = Arrays.asList(newVersion.split("\\."));
        List<String> oldVerList = Arrays.asList(olderVerison.split("\\."));

        int diff = newVerList.size() - oldVerList.size();
        List<String> newList = new ArrayList<>();
        if (diff > 0) {
            newList.addAll(oldVerList);
            for (int i = 0; i < diff; i++) {
                newList.add("0");
            }
            return examineArray(newList, newVerList, diff);
        } else if (diff < 0) {
            newList.addAll(newVerList);
            for (int i = 0; i < -diff; i++) {
                newList.add("0");
            }
            return examineArray(oldVerList, newList, diff);
        } else {
            return examineArray(oldVerList, newVerList, diff);
        }

    }

    public static boolean examineArray(List<String> oldList, List<String> newList, int diff) {
        boolean newVersionGreater = false;
        for (int i = 0; i < oldList.size(); i++) {
            if (Integer.parseInt(newList.get(i)) > Integer.parseInt(oldList.get(i))) {
                newVersionGreater = true;
                break;
            } else if (Integer.parseInt(newList.get(i)) < Integer.parseInt(oldList.get(i))) {
                newVersionGreater = false;
                break;
            } else {
                newVersionGreater = diff > 0;
            }
        }

        return newVersionGreater;
    }

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;
}

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

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);
    }
}

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!

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。