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

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
}

其他回答

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!

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;
}

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。

@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