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
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。