如何總結Java中的隨機字母數組？

Question

我想問一下是否有任何好的算法來總結隨機字符。

我期望在Cmd =上輸出

Random lowercase letters are  : 

m k o o b g q l y s z h h n j x b h l a
p a l x a v z e o w d b y x z o o v u g
o u l f k k w l s t o s u i r f k c e r
x i f x y o s e w f h j j c j q v i f g
q k z n o f j o a g i u t r v a l p o t

Iteration every characters : 

a : 5, b : 3, c : 2, d : 1, e : 3, f : 6, g : 4, i : 4, j : 5,
k : 5, l : 6, m : 1, n : 2, o : 11, p : 2, q : 3, r : 3, s : 4, t : 3,
u : 4, v : 4, w : 3, x : 5, y : 3, z : 4

這是我的源代碼=

public class CountRandomLettersArray {
/** Main method */
public static void main(String[] args) {
    //declare and create an array
    char[] kar = createArray();

    //show lowercase character array
    System.out.println("Random lowercase letters are  : ");
    System.out.println();
    showArray(kar);

    //counting every character loop
    int[] count = countLetter(kar);

//show sum of every character iteration
System.out.println();
System.out.println("Iteration every characters : ");
System.out.println();
showCount(count);
}

/** counting every character loop method */
public static int[] countLetter(char[] kar) {
    //declare and create an array that holds 26 of integer data type (26 means sum of 'a' to 'z')
    int[] count = new int[26];

    //count the loop for every lowercase letter in kar[] array
    for (int i = 0; i < kar.length; i++) {
        if (kar[i] == 'a')
            count[0]++;
        if (kar[i] == 'b')
            count[1]++;
        if (kar[i] == 'c')
            count[2]++;
        if (kar[i] == 'd')
            count[3]++;
        if (kar[i] == 'e')
            count[4]++;
        if (kar[i] == 'f')
            count[5]++;
        if (kar[i] == 'g')
            count[6]++;
        if (kar[i] == 'h')
            count[7]++;
        if (kar[i] == 'i')
            count[8]++;
        if (kar[i] == 'j')
            count[9]++;
        if (kar[i] == 'k')
            count[10]++;
        if (kar[i] == 'l')
            count[11]++;
        if (kar[i] == 'm')
            count[12]++;
        if (kar[i] == 'n')
            count[13]++;
        if (kar[i] == 'o')
            count[14]++;
        if (kar[i] == 'p')
            count[15]++;
        if (kar[i] == 'q')
            count[16]++;
        if (kar[i] == 'r')
            count[17]++;
        if (kar[i] == 's')
            count[18]++;
        if (kar[i] == 't')
            count[19]++;
        if (kar[i] == 'u')
            count[20]++;
        if (kar[i] == 'v')
            count[21]++;
        if (kar[i] == 'w')
            count[22]++;
        if (kar[i] == 'x')
            count[23]++;
        if (kar[i] == 'y')
            count[24]++;
        if (kar[i] == 'z')
            count[25]++;
    }

    return count;
}

/** generate a random array of lowercase letters method*/
public static char[] createArray () {
    //declare and create an array of 100 chars
    char[] kar = new char[100];

    //generate lowercase letters
    for (int i = 0; i < kar.length; i++) 
        kar[i] = RandomCharacters.getLowercase();

        return kar;
}

/** show characters array method*/
public static void showArray (char[] kar) {
    //shows five lines of lowercase letters
    final int CHARACTERS_PER_LINE = 20;
    for (int i = 0; i < kar.length; i++) {
        if ((i + 1) % CHARACTERS_PER_LINE == 0) 
            System.out.println(kar[i]);
        else
            System.out.print(kar[i] + " ");
    }
}


/** show sum of characters iteration method*/
public static void showCount(int[] count) {
final int SUM_OF_CHARACTERS_PER_LINE = 10;  
    for (int i = 0; i < count.length; i++) {
        if ((i + 1) % SUM_OF_CHARACTERS_PER_LINE == 0)
            System.out.println((char) (i + 'a') + " : " + count[i] + ", ");
        else
            System.out.print((char) (i + 'a') + " : " + count[i] + ", ");
    }
}
}

但這給了我這個錯誤： 編譯器錯誤

讓我感到困惑的是為什么為什么int無法在count []數組上轉換為int []（在第13行），而char卻可以適合kar []數組上的char []。 因此，kar []數組旨在僅顯示隨機的小寫字母，而count []數組旨在總結kar []數組具有的每個字符。

順便說一句，這是RandomCharacters.getLowercase（）->第93行的源代碼：

public class RandomCharacters {
/** Generate random characters from ch1 to ch2 */
public static char getRandomCharacters(char ch1, char ch2) {
    return (char) (ch1 + Math.random() * (ch2 - ch1 + 1));
}

/** Generate random lowercase letters (this is the method that I used) */
public static char getLowercase() {
    return getRandomCharacters('a', 'z');
}

/** Generate random uppercase letters */
public static char getUppercase() {
    return getRandomCharacters('A', 'Z');
}
}

我非常感謝您可以提供的任何幫助。 謝謝。

更新：問題已解決。 感謝Mohammed Housseyn Taleb。 為您+1。

Answer 1

試試這個代碼

    /**
     *
     * @author Taleb
     */
    public class Counter {
        static char[] letters;
        static int[] counts;
        public static void main(String[] args) {
            letters = new char[]{'a','a','a','a','a','a','b','b','c','c','c','d','d','d','d','d'
            ,'d','e','e','f','f','f'};
            counts = new int[]{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

            for (char letter : letters) {
                counts[letter-'a']++;
            }
            for (int i = 0; i < 26; i++) {
                char currentChar =(char) ('a'+i);
                System.out.println(currentChar+" = "+counts[i]);
            }

        }
    }

那么您可以使用此版本來處理A或a：

 letters = new char[]{'a','A','a','A','a','A','b','B','c','c','C','d','d','F','d','d'
            ,'d','e','E','f','F','f'};
            counts = new int[]{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

            for (char letter : letters) {
                final int index = letter-'A';
                counts[index % 32]++;
            }

Answer 2

嘗試這樣的事情：

public Map<Character, Integer> count( final char[] text )
{
    final Map<Character, Integer> result = new TreeMap<>();
    for ( final char c : text )
    {
        Integer i = result.get( c );
        if ( null == i )
            i = 0;
        result.put( c, i + 1 );
    }
    return result;
}

至於你的問題：

//counting every character loop
int[] count = countLetter(kar);

方法countLetter()返回單個int，並且無法將其轉換為int[] ...

Answer 3

您可以簡單地將char值用作數組索引：

int[] count = new int[...]; //size to support every letter

for(char letter : charArray) {
    count[letter]++;
}

它不是最快的，但是非常優雅。

---

要打印值和數量：

for(int i = 0; i < count.length; i++) {
    char letter = i;
    int amount = count[i];

    System.out.println("There are "+ count +" occurrences of "+ letter);
}

---

為了減小陣列大小，只需所有值在由97（數值為移位a ）：

int offset = 97; // 97

for(char letter : charArray) {
    count[letter - offset]++;
}

Answer 4

這是使用Java 8的解決方案：

char[] letters = {'a', 'a', 'a', 'a', 'a', 'a', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'd', 'd'
        , 'd', 'e', 'e', 'f', 'f', 'f'};

IntStream.range(0, letters.length)
        .mapToObj(i -> letters[i])
        .collect(Collectors.groupingBy(Function.identity(), TreeMap::new, Collectors.counting()))
        .forEach((character, count) -> System.out.println(character + " = " + count));