生成字母表中所有字母數組的更好方法

Question

現在我正在做

for (char c = 'a'; c <= 'z'; c++) {
    alphabet[c - 'a'] = c;
}

但是有更好的方法嗎？ 類似於 Scala 的'a' to 'z'

Answer 1

我認為這最終會更干凈一些，您不必處理減法和索引：

char[] alphabet = "abcdefghijklmnopqrstuvwxyz".toCharArray();

Answer 2

char[] alphabet = {'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'};

Answer 3

此getAlphabet方法使用與該問題類似的技術來為任意語言生成字母表。

將任何語言定義為枚舉，然后調用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);

結果

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;
    }
}

Answer 4

這是一個有趣的 Unicode 解決方案：

int charAmount = 'z' - 'a' + 1;

char[] alpha = new char[charAmount];
for(int i = 0; i < charAmount ; i++){
    alpha[i] = (char)('a' + i);
}

System.out.println(alpha); //abcdefghijklmnopqrstuvwxyz

這會生成字母表的小寫版本。
如果要大寫，可以在('a' + i)處將 'a' 替換為 'A' 。

Answer 5

在帶有 Stream API 的 Java 8 中，您可以執行此操作。

IntStream.rangeClosed('A', 'Z').mapToObj(var -> (char) var).forEach(System.out::println);

Answer 6

如果您使用的是 Java 8

char[] charArray = IntStream.rangeClosed('A', 'Z')
    .mapToObj(c -> "" + (char) c).collect(Collectors.joining()).toCharArray();

Answer 7

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"};

Answer 8

一旦我確定你會得到a到z字母，請檢查這個：

for (char c = 'a'; c <= 'z'; c++) {
    al.add(c);
}
System.out.println(al);'

Answer 9

使用io.vavr

public static char[] alphanumericAlphabet() {
    return CharSeq
            .rangeClosed('0','9')
            .appendAll(CharSeq.rangeClosed('a','z'))
            .appendAll(CharSeq.rangeClosed('A','Z'))
            .toCharArray();
}

Answer 10

以下是基於@tom thomas的回答的一些替代方案。

字符數組：

char[] list = IntStream.concat(
                IntStream.rangeClosed('0', '9'),
                IntStream.rangeClosed('A', 'Z')
        ).mapToObj(c -> (char) c+"").collect(Collectors.joining()).toCharArray();

字符串數組：

注意：如果您的分隔符也是值之一，則無法正常工作。

String[] list = IntStream.concat(
                IntStream.rangeClosed('0', '9'),
                IntStream.rangeClosed('A', 'Z')
        ).mapToObj(c -> (char) c+",").collect(Collectors.joining()).split(",");

字符串列表：

注意：如果您的分隔符也是值之一，則無法正常工作。

List<String> list = Arrays.asList(IntStream.concat(
                IntStream.rangeClosed('0', '9'),
                IntStream.rangeClosed('A', 'Z')
        ).mapToObj(c -> (char) c+",").collect(Collectors.joining()).split(","));

Answer 11

對於搜索 Kotlin 解決方案並在此處結束的 Android 開發人員：

// Creates List<Char>
val chars1 = ('a'..'z').toList()

// Creates Array<Char> (boxed)
val chars2 = ('a'..'z').toList().toTypedArray()

// Creates CharArray (unboxed)
val chars3 = CharArray(26) { 'a' + it }

// Creates CharArray (unboxed)
val chars4 = ('a'..'z').toArray()
fun CharRange.toArray() = CharArray(count()) { 'a' + it }

要了解我所說的“裝箱”和“未裝箱”的意思，請參閱這篇文章。
非常感謝這個 Kotlin 討論線程。

Answer 12

簡單是一種美德。 使用這個自然可讀的數組：

char alphabet[] = {'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'};

Answer 13

char[] abc = new char[26];

for(int i = 0; i<26;i++) {
    abc[i] = (char)('a'+i);
}

Answer 14

使用 Java 8 流

  char [] alphabets = Stream.iterate('a' , x -> (char)(x + 1))
            .limit(26)
            .map(c -> c.toString())
            .reduce("", (u , v) -> u + v).toCharArray();

Answer 15

要獲取除小寫字母之外的大寫字母，您還可以執行以下操作：

String alphabetWithUpper = "abcdefghijklmnopqrstuvwxyz" + "abcdefghijklmnopqrstuvwxyz".toUpperCase();
char[] letters = alphabetWithUpper.toCharArray();

Answer 16

在 kotlin 中你可以這樣做

var alphabet = "abcdefghijklmnopqrstuvwxyz".toCharArray()

Answer 17

var alphabet = IntStream.rangeClosed('a', 'z')
    .boxed()
    .map(Character::toChars)
    .map(String::valueOf)
    .toList();

筆記

• rangeClosed()以使其包含'z'
• boxed()以便從IntStream創建一個列表
• toList()創建一個不可修改的列表，但顯然僅從Java 16 開始可用

Answer 18

現在我在做

for (char c = 'a'; c <= 'z'; c++) {
    alphabet[c - 'a'] = c;
}

但是有沒有更好的方法來做到這一點？ 類似於 Scala 的'a' to 'z'

Answer 19

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));
    } 

Answer 20

for (char letter = 'a'; letter <= 'z'; letter++)
{
    System.out.println(letter);
}