现在我在做
for (char c = 'a'; c <= 'z'; c++) {
alphabet[c - 'a'] = c;
}
但是有更好的方法吗?类似于Scala的'a' to 'z'
现在我在做
for (char c = 'a'; c <= 'z'; c++) {
alphabet[c - 'a'] = c;
}
但是有更好的方法吗?类似于Scala的'a' to 'z'
"abcdefghijklmnopqrstuvwxyz".toCharArray()更好?
Answers:
我认为这样可以使操作更简洁,您不必处理减法和索引编制:
char[] alphabet = "abcdefghijklmnopqrstuvwxyz".toCharArray();
这是一个有趣的Unicode解决方案。
char[] alpha = new char[26]
for(int i = 0; i < 26; i++){
alpha[i] = (char)(97 + i)
}
这会生成小写字母的字母表,如果要大写,可以将“ 97”替换为“ 65”。
希望这可以帮助。
char保留一个UTF-16单位代码,其中一个或两个组成一个Unicode代码点。称它为ASCII会分散这些事实。
^或至少添加内容,我将对此表示支持~
k是多余的。做吧'a' + i。
在具有Stream API的Java 8中,您可以执行此操作。
IntStream.rangeClosed('A', 'Z').mapToObj(var -> (char) var).forEach(System.out::println);
将任何语言定义为枚举,并调用getAlphabet();
char[] armenianAlphabet = getAlphabet(LocaleLanguage.ARMENIAN);
char[] russianAlphabet = getAlphabet(LocaleLanguage.RUSSIAN);
// get uppercase alphabet
char[] currentAlphabet = getAlphabet(true);
System.out.println(armenianAlphabet);
System.out.println(russianAlphabet);
System.out.println(currentAlphabet);
结果
I / System.out:աբգդեզէըթժիլխծկհձղճմյնշոչպջռսվտրցւփքօֆ
I / System.out:абвгдежзийклмнопрстуфхцчшщъыьэюя
I / System.out:ABCDEFGHIJKLMNOPQRSTUVWXYZ
private char[] getAlphabet(){
return getAlphabet(false);
}
private char[] getAlphabet(boolean flagToUpperCase){
Locale locale = getResources().getConfiguration().locale;
LocaleLanguage language = LocaleLanguage.getLocalLanguage(locale);
return getAlphabet(language, flagToUpperCase);
}
private char[] getAlphabet(LocaleLanguage localeLanguage, boolean flagToUpperCase){
if (localeLanguage == null)
localeLanguage = LocaleLanguage.ENGLISH;
char firstLetter = localeLanguage.getFirstLetter();
char lastLetter = localeLanguage.getLastLetter();
int alphabetSize = lastLetter - firstLetter + 1;
char[] alphabet = new char[alphabetSize];
for (int index = 0; index < alphabetSize; index++ ){
alphabet[index] = (char) (index + firstLetter);
}
if (flagToUpperCase){
alphabet = new String(alphabet).toUpperCase().toCharArray();
}
return alphabet;
}
private enum LocaleLanguage{
ARMENIAN(new Locale("hy"), 'ա', 'ֆ'),
RUSSIAN(new Locale("ru"), 'а','я'),
ENGLISH(new Locale("en"), 'a','z');
private final Locale mLocale;
private final char mFirstLetter;
private final char mLastLetter;
LocaleLanguage(Locale locale, char firstLetter, char lastLetter) {
this.mLocale = locale;
this.mFirstLetter = firstLetter;
this.mLastLetter = lastLetter;
}
public Locale getLocale() {
return mLocale;
}
public char getFirstLetter() {
return mFirstLetter;
}
public char getLastLetter() {
return mLastLetter;
}
public String getDisplayLanguage(){
return getLocale().getDisplayLanguage();
}
public String getDisplayLanguage(LocaleLanguage locale){
return getLocale().getDisplayLanguage(locale.getLocale());
}
@Nullable
public static LocaleLanguage getLocalLanguage(Locale locale){
if (locale == null)
return LocaleLanguage.ENGLISH;
for (LocaleLanguage localeLanguage : LocaleLanguage.values()){
if (localeLanguage.getLocale().getLanguage().equals(locale.getLanguage()))
return localeLanguage;
}
return null;
}
}
如果您使用的是Java 8
char[] charArray = IntStream.rangeClosed('A', 'Z')
.mapToObj(c -> "" + (char) c).collect(Collectors.joining()).toCharArray();
Stream<Character> CharStream = IntStream.rangeClosed('a', 'z').mapToObj(c -> (char) c);并从那时开始使用它。stackoverflow.com/questions/22435833/...
char[]-为此,使用流无法避免装箱和拆箱。
一旦确定您可以a使用z字母,请检查一下:
for (char c = 'a'; c <= 'z'; c++) {
al.add(c);
}
System.out.println(al);'
static String[] AlphabetWithDigits = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"};
public static char[] alphanumericAlphabet() {
return CharSeq
.rangeClosed('0','9')
.appendAll(CharSeq.rangeClosed('a','z'))
.appendAll(CharSeq.rangeClosed('A','Z'))
.toCharArray();
}
根据汤姆·托马斯的回答,这里有一些替代方案。
char[] list = IntStream.concat(
IntStream.rangeClosed('0', '9'),
IntStream.rangeClosed('A', 'Z')
).mapToObj(c -> (char) c+"").collect(Collectors.joining()).toCharArray();
注意:如果您的定界符也是值之一,也将无法正常工作。
String[] list = IntStream.concat(
IntStream.rangeClosed('0', '9'),
IntStream.rangeClosed('A', 'Z')
).mapToObj(c -> (char) c+",").collect(Collectors.joining()).split(",");
注意:如果您的定界符也是值之一,也将无法正常工作。
List<String> list = Arrays.asList(IntStream.concat(
IntStream.rangeClosed('0', '9'),
IntStream.rangeClosed('A', 'Z')
).mapToObj(c -> (char) c+",").collect(Collectors.joining()).split(","));
最终,您将获得一个char array字母。为什么您会如此艰苦地使用loop?
只是
char[] alphabet=new char[]{'a','b',.........,'z'}
import java.util.*;
public class Experiments{
List uptoChar(int i){
char c='a';
List list = new LinkedList();
for(;;) {
list.add(c);
if(list.size()==i){
break;
}
c++;
}
return list;
}
public static void main (String [] args) {
Experiments experiments = new Experiments();
System.out.println(experiments.uptoChar(26));
}
for(;;)与外部计数器和break...并且没有检查i>='a'