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
基于https://stackoverflow.com/a/27891752/2642478
class Version(private val value: String) : Comparable<Version> {
private val splitted by lazy { value.split("-").first().split(".").map { it.toIntOrNull() ?: 0 } }
override fun compareTo(other: Version): Int {
for (i in 0 until maxOf(splitted.size, other.splitted.size)) {
val compare = splitted.getOrElse(i) { 0 }.compareTo(other.splitted.getOrElse(i) { 0 })
if (compare != 0)
return compare
}
return 0
}
}
你可以用like:
System.err.println(Version("1.0").compareTo( Version("1.0")))
System.err.println(Version("1.0") < Version("1.1"))
System.err.println(Version("1.10") > Version("1.9"))
System.err.println(Version("1.10.1") > Version("1.10"))
System.err.println(Version("0.0.1") < Version("1"))
其他回答
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;
}
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
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!
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;
}