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
public class VersionComparator {
/* loop through both version strings
* then loop through the inner string to computer the val of the int
* for each integer read, do num*10+<integer read>
* and stop when stumbling upon '.'
* When '.' is encountered...
* see if '.' is encountered for both strings
* if it is then compare num1 and num2
* if num1 == num2... iterate over p1++, p2++
* else return (num1 > num2) ? 1 : -1
* If both the string end then compare(num1, num2) return 0, 1, -1
* else loop through the longer string and
* verify if it only has trailing zeros
* If it only has trailing zeros then return 0
* else it is greater than the other string
*/
public static int compareVersions(String v1, String v2) {
int num1 = 0;
int num2 = 0;
int p1 = 0;
int p2 = 0;
while (p1 < v1.length() && p2 < v2.length()) {
num1 = Integer.parseInt(v1.charAt(p1) + "");
num2 = Integer.parseInt(v2.charAt(p2) + "");
p1++;
p2++;
while (p1 < v1.length() && p2 < v2.length() && v1.charAt(p1) != '.' && v2.charAt(p2) != '.') {
if (p1 < v1.length()) num1 = num1 * 10 + Integer.parseInt(v1.charAt(p1) + "");
if (p2 < v2.length()) num2 = num2 * 10 + Integer.parseInt(v2.charAt(p2) + "");
p1++;
p2++;
}
if (p1 < v1.length() && p2 < v2.length() && v1.charAt(p1) == '.' && v2.charAt(p2) == '.') {
if ((num1 ^ num2) == 0) {
p1++;
p2++;
}
else return (num1 > num2) ? 1 : -1;
}
else if (p1 < v1.length() && p2 < v2.length() && v1.charAt(p1) == '.') return -1;
else if (p1 < v1.length() && p2 < v2.length() && v2.charAt(p2) == '.') return 1;
}
if (p1 == v1.length() && p2 == v2.length()) {
if ((num1 ^ num2) == 0) return 0;
else return (num1 > num2) ? 1 : -1;
}
else if (p1 == v1.length()) {
if ((num1 ^ num2) == 0) {
while (p2 < v2.length()) {
if (v2.charAt(p2) != '.' && v2.charAt(p2) != '0') return -1;
p2++;
}
return 0;
}
else return (num1 > num2) ? 1 : -1;
}
else {
if ((num1 ^ num2) == 0) {
while (p1 < v1.length()) {
if (v1.charAt(p1) != '.' && v1.charAt(p1) != '0') return 1;
p1++;
}
return 0;
}
else return (num1 > num2) ? 1 : -1;
}
}
public static void main(String[] args) {
System.out.println(compareVersions("11.23", "11.21.1.0.0.1.0") ^ 1);
System.out.println(compareVersions("11.21.1.0.0.1.0", "11.23") ^ -1);
System.out.println(compareVersions("11.23", "11.23.0.0.0.1.0") ^ -1);
System.out.println(compareVersions("11.2", "11.23") ^ -1);
System.out.println(compareVersions("11.23", "11.21.1.0.0.1.0") ^ 1);
System.out.println(compareVersions("1.21.1.0.0.1.0", "2.23") ^ -1);
System.out.println(compareVersions("11.23", "11.21.1.0.0.1.0") ^ 1);
System.out.println(compareVersions("11.23.0.0.0.0.0", "11.23") ^ 0);
System.out.println(compareVersions("11.23", "11.21.1.0.0.1.0") ^ 1);
System.out.println(compareVersions("1.5.1.3", "1.5.1.3.0") ^ 0);
System.out.println(compareVersions("1.5.1.4", "1.5.1.3.0") ^ 1);
System.out.println(compareVersions("1.2.1.3", "1.5.1.3.0") ^ -1);
System.out.println(compareVersions("1.2.1.3", "1.22.1.3.0") ^ -1);
System.out.println(compareVersions("1.222.1.3", "1.22.1.3.0") ^ 1);
}
}
其他回答
public static void main(String[] args) {
String version1 = "1.0";
String version2 = "1.0.0";
String[] version1_splits = version1.split("\\.");
String[] version2_splits = version2.split("\\.");
int length = version1_splits.length >= version2_splits.length ? version1_splits.length : version2_splits.length;
int i=0;
for(;i<length;i++){
int version1_int = getValue(version1_splits,i);
int version2_int = getValue(version2_splits,i);
if(version1_int > version2_int){
System.out.println("version1 > version2");
break;
}
else if(version1_int < version2_int){
System.out.println("version2 > version1");
break;
}
else{
if(i == length-1)
System.out.println("version1 = version2");
}
}
}
private static int getValue(String[] version1_splits, int i) {
int temp;
try{
temp = Integer.valueOf(version1_splits[i]);
}
catch(IndexOutOfBoundsException e){
temp=0;
}
return temp;
}
使用Java 9自带的Version类
import java.util.*;
import java.lang.module.ModuleDescriptor.Version;
class Main {
public static void main(String[] args) {
var versions = Arrays.asList(
"1.0.2",
"1.0.0-beta.2",
"1.0.0",
"1.0.0-beta",
"1.0.0-alpha.12",
"1.0.0-beta.11",
"1.0.1",
"1.0.11",
"1.0.0-rc.1",
"1.0.0-alpha.1",
"1.1.0",
"1.0.0-alpha.beta",
"1.11.0",
"1.0.0-alpha.12.ab-c",
"0.0.1",
"1.2.1",
"1.0.0-alpha",
"1.0.0.1", // Also works with a number of sections different than 3
"1.0.0.2",
"2",
"10",
"1.0.0.10"
);
versions.stream()
.map(Version::parse)
.sorted()
.forEach(System.out::println);
}
}
在网上试试!
输出:
0.0.1
1.0.0-alpha
1.0.0-alpha.1
1.0.0-alpha.12
1.0.0-alpha.12.ab-c
1.0.0-alpha.beta
1.0.0-beta
1.0.0-beta.2
1.0.0-beta.11
1.0.0-rc.1
1.0.0
1.0.0.1
1.0.0.2
1.0.0.10
1.0.1
1.0.2
1.0.11
1.1.0
1.2.1
1.11.0
2
10
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。
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!
如果你的项目中已经有Jackson,你可以使用com.fasterxml.jackson.core.Version:
import com.fasterxml.jackson.core.Version;
import org.junit.Test;
import static org.junit.Assert.assertTrue;
public class VersionTest {
@Test
public void shouldCompareVersion() {
Version version1 = new Version(1, 11, 1, null, null, null);
Version version2 = new Version(1, 12, 1, null, null, null);
assertTrue(version1.compareTo(version2) < 0);
}
}