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
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!
其他回答
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
}
我写了一个小的Java/Android库来比较版本号:https://github.com/G00fY2/version-compare
它的基本功能是:
public int compareVersions(String versionA, String versionB) {
String[] versionTokensA = versionA.split("\\.");
String[] versionTokensB = versionB.split("\\.");
List<Integer> versionNumbersA = new ArrayList<>();
List<Integer> versionNumbersB = new ArrayList<>();
for (String versionToken : versionTokensA) {
versionNumbersA.add(Integer.parseInt(versionToken));
}
for (String versionToken : versionTokensB) {
versionNumbersB.add(Integer.parseInt(versionToken));
}
final int versionASize = versionNumbersA.size();
final int versionBSize = versionNumbersB.size();
int maxSize = Math.max(versionASize, versionBSize);
for (int i = 0; i < maxSize; i++) {
if ((i < versionASize ? versionNumbersA.get(i) : 0) > (i < versionBSize ? versionNumbersB.get(i) : 0)) {
return 1;
} else if ((i < versionASize ? versionNumbersA.get(i) : 0) < (i < versionBSize ? versionNumbersB.get(i) : 0)) {
return -1;
}
}
return 0;
}
这个代码片段不提供任何错误检查或处理。除此之外，我的库还支持像“1.2-rc”>“1.2-beta”这样的后缀。
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!
这篇旧文章的另一个解决方案(对那些可能有帮助的人来说):
public class Version implements Comparable<Version> {
private String version;
public final String get() {
return this.version;
}
public Version(String version) {
if(version == null)
throw new IllegalArgumentException("Version can not be null");
if(!version.matches("[0-9]+(\\.[0-9]+)*"))
throw new IllegalArgumentException("Invalid version format");
this.version = version;
}
@Override public int compareTo(Version that) {
if(that == null)
return 1;
String[] thisParts = this.get().split("\\.");
String[] thatParts = that.get().split("\\.");
int length = Math.max(thisParts.length, thatParts.length);
for(int i = 0; i < length; i++) {
int thisPart = i < thisParts.length ?
Integer.parseInt(thisParts[i]) : 0;
int thatPart = i < thatParts.length ?
Integer.parseInt(thatParts[i]) : 0;
if(thisPart < thatPart)
return -1;
if(thisPart > thatPart)
return 1;
}
return 0;
}
@Override public boolean equals(Object that) {
if(this == that)
return true;
if(that == null)
return false;
if(this.getClass() != that.getClass())
return false;
return this.compareTo((Version) that) == 0;
}
}
Version a = new Version("1.1");
Version b = new Version("1.1.1");
a.compareTo(b) // return -1 (a<b)
a.equals(b) // return false
Version a = new Version("2.0");
Version b = new Version("1.9.9");
a.compareTo(b) // return 1 (a>b)
a.equals(b) // return false
Version a = new Version("1.0");
Version b = new Version("1");
a.compareTo(b) // return 0 (a=b)
a.equals(b) // return true
Version a = new Version("1");
Version b = null;
a.compareTo(b) // return 1 (a>b)
a.equals(b) // return false
List<Version> versions = new ArrayList<Version>();
versions.add(new Version("2"));
versions.add(new Version("1.0.5"));
versions.add(new Version("1.01.0"));
versions.add(new Version("1.00.1"));
Collections.min(versions).get() // return min version
Collections.max(versions).get() // return max version
// WARNING
Version a = new Version("2.06");
Version b = new Version("2.060");
a.equals(b) // return false
编辑:
@daiscog:谢谢你的评论，这段代码是为Android平台开发的，由谷歌推荐，方法“匹配”检查整个字符串，不像Java使用监管模式。(Android文档- JAVA文档)
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;
}
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