如何編輯此編碼以顯示頻率為0的字符。例如：a=0。因為現在它只顯示freq>0的字符

撰寫一個程式，讀取一行文本String并顯示文本中出現的所有字母，每行一個，按字母順序，以及每個字母在文本中出現的次數。

為此，您必須使用int長度型別的陣列26，以便索引處的元素0包含a' 的數量，索引處的元素1包含b' 的數量，依此類推。

允許大寫和小寫字母作為輸入，但將相同字母的大寫和小寫版本視為相同。

提示：使用方法chatAt(int index)的String類指定索引處得到的字串中的個性。

public class Test {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int[] letters = new int[26];
        char choice;

        while (true) {
            // taking user input
            System.out.println("Please enter text ending with period:");
            String text = sc.nextLine();

            // converting it lowercase
            text = getActualText(text).toLowerCase();

            char c = 'a';

            for (int i = 0; i < letters.length; i  )
                // increasing character by 1
                letters[i] = countLetters(text, c  );

            System.out.println("\nThe frequency of the letters");
            c = 'a';

            for (int i = 0; i < letters.length; i  ) {
                // showing only those letters whose frequnecy is greater than 0
                if (letters[i] != 0)
                    System.out.println(c   ": "   letters[i]);

                c  ;
            }

            System.out.print("Would you like to try another text?(Y/N) ");
            choice = sc.nextLine().charAt(0);

            if (choice == 'n' || choice == 'N')
                break;
        }
    }

    private static int countLetters(String text, char c) {
        int count = 0;

        for (int i = 0; i < text.length(); i  )
            // counting the frequency
            if (text.charAt(i) == c)
                count  ;

        return count;
    }

    /**
     * This method will extract the first sentence from a text ending with full stop(.)
     */
    private static String getActualText(String text) {
        String newText = "";

        for (int i = 0; i < text.length(); i  ) {
            if (text.charAt(i) == '.')
                // breaking out of the loop if the full stop is found
                break;

            // adding it to the text
            newText  = text.charAt(i)   "";
        }

        return newText;
    }

}

uj5u.com熱心網友回復：

嘗試將現有條件更改為以下新條件：

現有條件：（允許不等于0的頻率）：

if(letters[i] != 0) {//showing only those letters whose frequency is greater than 0

新條件：（允許大于或等于0的頻率）：

if(letters[i] >= 0) {

uj5u.com熱心網友回復：

遍歷文本一次并計算每個字母的出現次數就足夠了。然后只顯示帶有 count 的字母>0。

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);

    do {
        System.out.print("\nEnter the text: ");
        String str = scan.nextLine();
        print(histogram(str));
    } while (shouldContinue(scan));
}

private static int[] histogram(String str) {
    int[] letters = new int[26];

    for (int i = 0; i < str.length(); i  )
        if (Character.isLetter(str.charAt(i)))
            letters[Character.toLowerCase(str.charAt(i)) - 'a']  ;

    return letters;
}

private static void print(int[] letters) {
    System.out.println("The frequency of the letters:");

    for (int i = 0; i < letters.length; i  )
        if (letters[i] > 0)
            System.out.println((char)('a'   i)   ": "   letters[i]);
}

private static boolean shouldContinue(Scanner scan) {
    while (true) {
        System.out.print("Would you like to try another text (Y/N)? ");
        String str = scan.nextLine();

        if (str.length() != 1)
            continue;
        if ("Y".equalsIgnoreCase(str))
            return true;
        if ("N".equalsIgnoreCase(str))
            return false;
    }
}

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