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 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;
}

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

想知道为什么每个人都假设版本只由整数组成-在我的情况下，它不是。

为什么要重新发明轮子(假设版本遵循Semver标准)

首先通过Maven安装https://github.com/vdurmont/semver4j

然后使用这个库

Semver sem = new Semver("1.2.3");
sem.isGreaterThan("1.2.2"); // true

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!

@alex在Kotlin上的帖子

class Version(inputVersion: String) : Comparable<Version> {

        var version: String
            private set

        override fun compareTo(other: Version) =
            (split() to other.split()).let {(thisParts, thatParts)->
                val length = max(thisParts.size, thatParts.size)
                for (i in 0 until length) {
                    val thisPart = if (i < thisParts.size) thisParts[i].toInt() else 0
                    val thatPart = if (i < thatParts.size) thatParts[i].toInt() else 0
                    if (thisPart < thatPart) return -1
                    if (thisPart > thatPart) return 1
                }
                 0
            }

        init {
            require(inputVersion.matches("[0-9]+(\\.[0-9]+)*".toRegex())) { "Invalid version format" }
            version = inputVersion
        }
    }

    fun Version.split() = version.split(".").toTypedArray()

用法:

Version("1.2.4").compareTo(Version("0.0.5")) //return 1