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
// VersionComparator.java
import java.util.Comparator;
public class VersionComparator implements Comparator {
public boolean equals(Object o1, Object o2) {
return compare(o1, o2) == 0;
}
public int compare(Object o1, Object o2) {
String version1 = (String) o1;
String version2 = (String) o2;
VersionTokenizer tokenizer1 = new VersionTokenizer(version1);
VersionTokenizer tokenizer2 = new VersionTokenizer(version2);
int number1 = 0, number2 = 0;
String suffix1 = "", suffix2 = "";
while (tokenizer1.MoveNext()) {
if (!tokenizer2.MoveNext()) {
do {
number1 = tokenizer1.getNumber();
suffix1 = tokenizer1.getSuffix();
if (number1 != 0 || suffix1.length() != 0) {
// Version one is longer than number two, and non-zero
return 1;
}
}
while (tokenizer1.MoveNext());
// Version one is longer than version two, but zero
return 0;
}
number1 = tokenizer1.getNumber();
suffix1 = tokenizer1.getSuffix();
number2 = tokenizer2.getNumber();
suffix2 = tokenizer2.getSuffix();
if (number1 < number2) {
// Number one is less than number two
return -1;
}
if (number1 > number2) {
// Number one is greater than number two
return 1;
}
boolean empty1 = suffix1.length() == 0;
boolean empty2 = suffix2.length() == 0;
if (empty1 && empty2) continue; // No suffixes
if (empty1) return 1; // First suffix is empty (1.2 > 1.2b)
if (empty2) return -1; // Second suffix is empty (1.2a < 1.2)
// Lexical comparison of suffixes
int result = suffix1.compareTo(suffix2);
if (result != 0) return result;
}
if (tokenizer2.MoveNext()) {
do {
number2 = tokenizer2.getNumber();
suffix2 = tokenizer2.getSuffix();
if (number2 != 0 || suffix2.length() != 0) {
// Version one is longer than version two, and non-zero
return -1;
}
}
while (tokenizer2.MoveNext());
// Version two is longer than version one, but zero
return 0;
}
return 0;
}
}
// VersionTokenizer.java
public class VersionTokenizer {
private final String _versionString;
private final int _length;
private int _position;
private int _number;
private String _suffix;
private boolean _hasValue;
public int getNumber() {
return _number;
}
public String getSuffix() {
return _suffix;
}
public boolean hasValue() {
return _hasValue;
}
public VersionTokenizer(String versionString) {
if (versionString == null)
throw new IllegalArgumentException("versionString is null");
_versionString = versionString;
_length = versionString.length();
}
public boolean MoveNext() {
_number = 0;
_suffix = "";
_hasValue = false;
// No more characters
if (_position >= _length)
return false;
_hasValue = true;
while (_position < _length) {
char c = _versionString.charAt(_position);
if (c < '0' || c > '9') break;
_number = _number * 10 + (c - '0');
_position++;
}
int suffixStart = _position;
while (_position < _length) {
char c = _versionString.charAt(_position);
if (c == '.') break;
_position++;
}
_suffix = _versionString.substring(suffixStart, _position);
if (_position < _length) _position++;
return true;
}
}
例子:
public class Main
{
private static VersionComparator cmp;
public static void main (String[] args)
{
cmp = new VersionComparator();
Test(new String[]{"1.1.2", "1.2", "1.2.0", "1.2.1", "1.12"});
Test(new String[]{"1.3", "1.3a", "1.3b", "1.3-SNAPSHOT"});
}
private static void Test(String[] versions) {
for (int i = 0; i < versions.length; i++) {
for (int j = i; j < versions.length; j++) {
Test(versions[i], versions[j]);
}
}
}
private static void Test(String v1, String v2) {
int result = cmp.compare(v1, v2);
String op = "==";
if (result < 0) op = "<";
if (result > 0) op = ">";
System.out.printf("%s %s %s\n", v1, op, v2);
}
}
输出:
1.1.2 == 1.1.2 ---> same length and value
1.1.2 < 1.2 ---> first number (1) less than second number (2) => -1
1.1.2 < 1.2.0 ---> first number (1) less than second number (2) => -1
1.1.2 < 1.2.1 ---> first number (1) less than second number (2) => -1
1.1.2 < 1.12 ---> first number (1) less than second number (12) => -1
1.2 == 1.2 ---> same length and value
1.2 == 1.2.0 ---> first shorter than second, but zero
1.2 < 1.2.1 ---> first shorter than second, and non-zero
1.2 < 1.12 ---> first number (2) less than second number (12) => -1
1.2.0 == 1.2.0 ---> same length and value
1.2.0 < 1.2.1 ---> first number (0) less than second number (1) => -1
1.2.0 < 1.12 ---> first number (2) less than second number (12) => -1
1.2.1 == 1.2.1 ---> same length and value
1.2.1 < 1.12 ---> first number (2) less than second number (12) => -1
1.12 == 1.12 ---> same length and value
1.3 == 1.3 ---> same length and value
1.3 > 1.3a ---> first suffix ('') is empty, but not second ('a') => 1
1.3 > 1.3b ---> first suffix ('') is empty, but not second ('b') => 1
1.3 > 1.3-SNAPSHOT ---> first suffix ('') is empty, but not second ('-SNAPSHOT') => 1
1.3a == 1.3a ---> same length and value
1.3a < 1.3b ---> first suffix ('a') compared to second suffix ('b') => -1
1.3a < 1.3-SNAPSHOT ---> first suffix ('a') compared to second suffix ('-SNAPSHOT') => -1
1.3b == 1.3b ---> same length and value
1.3b < 1.3-SNAPSHOT ---> first suffix ('b') compared to second suffix ('-SNAPSHOT') => -1
1.3-SNAPSHOT == 1.3-SNAPSHOT ---> same length and value
其他回答
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。
对于我的项目,我使用我的公共版本库https://github.com/raydac/commons-version 它包含两个辅助类-用于解析版本(解析后的版本可以与另一个版本对象进行比较,因为它是可比的)和VersionValidator,它允许检查一些表达式的版本,如!=ide-1.1.1,>idea-1.3.4-SNAPSHOT;<1.2.3
我写了一个名为MgntUtils的开源库,它有一个用于字符串版本的实用程序。它正确地比较它们,适用于版本范围等等。下面是这个库javadoc参见方法TextUtils.comapreVersions(…)它已经被大量使用并经过了良好的测试。下面这篇文章描述了这个库以及如何获取它。它可以作为Maven工件和在github上获得(包括源代码和JavaDoc)
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!
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'