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

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。

其他回答

@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

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

也许有人会对我的解决方案感兴趣:

class Version private constructor(private val versionString: String) : Comparable<Version> {

    private val major: Int by lazy { versionString.split(".")[0].toInt() }

    private val minor: Int by lazy { versionString.split(".")[1].toInt() }

    private val patch: Int by lazy {
        val splitArray = versionString.split(".")

        if (splitArray.size == 3)
            splitArray[2].toInt()
        else
            0
    }

    override fun compareTo(other: Version): Int {
        return when {
            major > other.major -> 1
            major < other.major -> -1
            minor > other.minor -> 1
            minor < other.minor -> -1
            patch > other.patch -> 1
            patch < other.patch -> -1
            else -> 0
        }
    }

    override fun equals(other: Any?): Boolean {
        if (other == null || other !is Version) return false
        return compareTo(other) == 0
    }

    override fun hashCode(): Int {
        return major * minor * patch
    }

    companion object {
        private fun doesContainsVersion(string: String): Boolean {
            val versionArray = string.split(".")

            return versionArray.size in 2..3
                    && versionArray[0].toIntOrNull() != null
                    && versionArray[1].toIntOrNull() != null
                    && (versionArray.size == 2 || versionArray[2].toIntOrNull() != null)
        }

        fun from(string: String): Version? {
            return if (doesContainsVersion(string)) {
                Version(string)
            } else {
                null
            }
        }
    }
}

用法:

val version1 = Version.from("3.2")
val version2 = Version.from("3.2.1")
version1 <= version2

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!

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。