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
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;
}
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
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
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;
}
您需要规范化版本字符串,以便对它们进行比较。类似的
import java.util.regex.Pattern;
public class Main {
public static void main(String... args) {
compare("1.0", "1.1");
compare("1.0.1", "1.1");
compare("1.9", "1.10");
compare("1.a", "1.9");
}
private static void compare(String v1, String v2) {
String s1 = normalisedVersion(v1);
String s2 = normalisedVersion(v2);
int cmp = s1.compareTo(s2);
String cmpStr = cmp < 0 ? "<" : cmp > 0 ? ">" : "==";
System.out.printf("'%s' %s '%s'%n", v1, cmpStr, v2);
}
public static String normalisedVersion(String version) {
return normalisedVersion(version, ".", 4);
}
public static String normalisedVersion(String version, String sep, int maxWidth) {
String[] split = Pattern.compile(sep, Pattern.LITERAL).split(version);
StringBuilder sb = new StringBuilder();
for (String s : split) {
sb.append(String.format("%" + maxWidth + 's', s));
}
return sb.toString();
}
}
打印
'1.0' < '1.1' '1.0.1' < '1.1' '1.9' < '1.10' “1。A ' > '1.9'