如何检测程序中的元音和辅音

Question

    String text;
    System.out.print ("Enter a String:");
    text = console.nextLine();

    int spaces = 0;
    int consonants = 0;
    int vowelcount = 0 ;

    for (int index = 0; index < text.length(); index++) {
    char letters = text.charAt(index);


    if (letters == 'A' || letters == 'a')
        vowelcount++;



    else if (letters != 'a' && letters != 'e' && letters != 'i' && letters != 'o' && letters != 'u')
        consonants++;

    }


        System.out.println ("Vowels:" + vowelcount  + "\nConsonants :" + consonants + "\nSpaces : " + spaces);

示例输出 字符串：汉娜 输出的最后一部分 检测到的元音：a 检测到辅音：h n n h

Answer 1

以下是一些帮助方法

public static boolean isVowel(char c){
    String vowels = "aeiouAEIOU";
    return vowels.contains(c);
}

public static boolean isConsanant(char c){
    String cons = "bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
    return cons.contains(c);
}

在这里使用

char c = line.charAt(i);
int vowelCount = 0;
int consanantCount = 0;
int space = 0;
int punctuation = 0;

if (isVowel(c))
    vowelCount++;
else if (isConsanant(c))
    consanantCount++;
else if (Character.isWhitepace(c))
    space++;
else
    punctuation++;

Answer 2

只需使用正则表达式，它只需要一行计数：

int spaces = text.replaceAll("\\S", "").length();
int consonants = text.replaceAll("(?i)[\\saeiou]", "").length();
int vowelcount = text.replaceAll("(?i)[^aeiou]", "").length();

这些都替换了匹配目标字符类型的字符而不是 - 有效删除它们 - 然后使用String.length（）为您提供计数。

Answer 3

您可以使用正则表达式：

• 空格：（？\ s +）
• 元音:(？[aeijo] +）

我想下一个应该很容易;）

然后使用组匹配功能并计算所有实例

（我大多数时候在构建正则表达式时使用正则表达式工具，例如http://gskinner.com/RegExr/）

Answer 4

要检测元音和辅音，您需要一个CONSONANTS字符数组，然后检查字符是否在此数组中。在这里你可以看到一个工作的例子，它计算辅音，元音和空格： import java.io.Console;

public class Vowels
{
    public static final char[] CONSONANTS =
    {
        'b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'y', 'z'
    };

    public static final char SPACE = ' ';

    public static char[] getConsonants()
    {
        return CONSONANTS;
    }

    public static boolean isConsonant(char c)
    {
        boolean isConsonant = false;
        for (int i = 0; i < getConsonants().length; i++)
        {
            if (getConsonants()[i] == c)
            {
                isConsonant = true;
                break;
            }
        }
        return isConsonant;
    }

    public static boolean isSpace(char c)
    {
        return SPACE == c;
    }

    public static void main(String[] args)
    {
        int spaces     = 0;
        int consonants = 0;
        int vowelcount = 0;

        Console console = System.console();
        console.format("Enter a String:");

        String text = console.readLine();;

        for (int index = 0; index < text.length(); index++)
        {
            char letter = text.charAt(index);
            if (!isSpace(letter))
            {
                if (isConsonant(letter))
                {
                    consonants++;
                }
                else
                {
                    vowelcount++;
                }
            }
            else
            {
                spaces++;
            }
        }

        System.out.println("Vowels:" + vowelcount + "\nConsonants :" + consonants + "\nSpaces : " + spaces);
    }
}

Answer 5

这是一种简单的方法，从How to count vowels and consonants

重新发布我的答案

public static void checkVowels(String s){
    System.out.println("Vowel Count: " + (s.length() - s.toLowerCase().replaceAll("a|e|i|o|u|", "").length()));
    //Also eliminating spaces, if any for the consonant count
    System.out.println("Consonant Count: " + (s.toLowerCase().replaceAll("a|e|i|o| |u", "").length()));
}

Answer 6

使用LinkedHashSet，因为它保留了顺序，并且不允许重复

//Check for vowel
public static boolean isVovel(char c) {

    if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
        return true;
    }
    return false;
}

public static void main(String[] args) {
    String input = "shashi is a good boy";

    char inter;
    String vov = "";
    String con = "";
    String inp;
    int len = input.length();

    LinkedHashSet<String> vovels = new LinkedHashSet<String>();
    LinkedHashSet<String> consonents = new LinkedHashSet<String>();

    for (int i = 0; i < len; i++) {
        inter = input.charAt(i);
        inp = Character.toString(inter);

        if (isVovel(inter)) {
            vov = Character.toString(inter);
            vovels.add(vov);
        } 
        else {
            con = Character.toString(inter);
            consonents.add(con);
        }
    }
    Iterator<String> it = consonents.iterator();

    while (it.hasNext()) {
        String value = it.next();
        if (" ".equals(value)) {
            it.remove();
        }
    }
    System.out.println(vovels);
    System.out.println(consonents);
}

Answer 7

此示例是一个类，该类从文件中以字符串形式读取文本，将该文本存储为char数组，并初始化该数组的每个元素，从而将该元素转换为String并查看其是否与辅音正则表达式或元音匹配正则表达式。它会根据正则表达式匹配项来增加consonantCount或vowelCount，最后打印出计数。

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.regex.Pattern;

public class CountVowlesAndConsonants {

public static void main (String [] args) throws IOException{
    CountVowlesAndConsonants countVowlesAndConsonants = new CountVowlesAndConsonants();
    countVowlesAndConsonants.countConsonatsAndVowles("/Users/johndoe/file.txt");

}
public void countConsonatsAndVowles(String file) throws IOException {
     String text = readFile(file);
     Pattern c = Pattern.compile("^(?![aeiouy]+)([a-z]+)$");
     Pattern v = Pattern.compile("^[aeiouy]+$");

     int vowelCount = 0;
     int consonantCount = 0;

     char [] textArray = text.toLowerCase().toCharArray();
     for( char textArraz : textArray ){
         String s = String.valueOf(textArraz);
         if(c.matcher(s).matches()) {
             consonantCount++;
         } else if (v.matcher(s).matches()) {
             vowelCount++;
         }
     }

     System.out.println("VowelCount is " + vowelCount + " Constant Count " + consonantCount);
}

public String readFile(String file) throws IOException {
    BufferedReader reader = new BufferedReader(new FileReader (file));
    String         line = null;
    StringBuilder  stringBuilder = new StringBuilder();

    try {
        while((line = reader.readLine()) != null) {
            stringBuilder.append(line);
        }
        return stringBuilder.toString();
    } finally {
        reader.close();
    }
}
}

Answer 8

我需要使用'loops for'来完成此功能，请遵循JavaScript中的示例：

#include