Question

我正在编写一个程序，用于计算用户输入的句子中的元音和辅音数量。我的下面的代码计算元音的数量，但它给出了辅音计数的奇怪数字。例如，如果我输入＆＃34; g＆＃34;我的辅音数为10。

import java.util.Scanner;

public class VowelCount{
  public static void main(String[] args){
    Scanner scan = new Scanner(System.in);
    System.out.print("Enter a sentence :");
    String sentence = scan.nextLine();

    String vowels = "aeiouAEIOU";
    int vowelCount = 0;
    int consCount = 0;
    int i;

      for(i = 0; i < sentence.length(); i += 1){
        char currentChar = sentence.charAt(i);
        int index;
        for(index = 0; index < vowels.length(); index += 1){
          if(vowels.charAt(index) == (currentChar)){
            vowelCount++;
          }else if(Character.isLetter(currentChar) && (vowels.charAt(index) == (currentChar))){
            consCount++;
          }
        }
      }
      System.out.println(consCount);
      System.out.println(vowelCount);
  }   
}

Answer 1

这很有效。我也对其进行了改进，请参阅vowels.indexOf（而不是手动迭代）和isLetter（校正错误的元音检查）和输出（添加;）和i++的行

public class VowelCount{
  public static void main(String[] args){
    Scanner scan = new Scanner(System.in);
    System.out.print("Enter a sentence :");
    String sentence = scan.nextLine();

    String vowels = "aeiouAEIOU";
    int vowelCount = 0;
    int consCount = 0;
    int i;

    int length = sentence.length();
    for(i = 0; i < length; i ++){
        char currentChar = sentence.charAt(i);
        if (vowels.indexOf(currentChar)>=0)
           vowelCount++;
        else if(Character.isLetter(currentChar))
                consCount++;
    }
    System.out.print(consCount+";");
    System.out.print(vowelCount);
  }
}

Answer 2

这是最终的解决方案：）

import java.util.Scanner;

public class VowelConsCount
{ public static void main(String args[])
    { System.out.print("Enter a sentence :");
      Scanner in = new Scanner(System.in);
      String sentence = in.nextLine();

      String vowels = "aeiouAEIOU";
      int vowelCount = 0;
      int consCount = 0;
      int i;

      for(i = 0; i < sentence.length(); i ++)
         { char currentChar = sentence.charAt(i);
           if (vowels.indexOf(currentChar)>=0)
                vowelCount++;
           else if(Character.isLetter(currentChar))
                consCount++;
         }
      System.out.print("\nNumber of vowels is "+vowelCount);
      System.out.print("\nNumber of consonant is "+consCount);
    }
}

Answer 3

在那一行： else if(Character.isLetter(currentChar) && (vowels.charAt(index) == (currentChar))){ 你正在检查你的字母是否是一封信，以及它是否是一个元音。您可以在条件测试中尝试：

else if(Character.isLetter(currentChar)){

Answer 4

问题在于你的内部for循环的其他部分。

boolean isVowel(char c){
   String vowels = "aeiouAEIOU";
   for(int index = 0; index < vowels.length(); index += 1)
   if(vowels.charAt(index) == c)
      return true;
   return false;
}

现在，在你获得currentChar后，试试这个

if(isVowel(currentChar))
   vowelCount++;
else
   constCount++;

Answer 5

您的代码有几个问题：

您使用System.out.print代替System.out.println，这会使阅读号码误导。

如果找到元音，您正在循环可能的元音并递增vowelCount但是如果没有，则递增consCount！。你应该分开两个：先检查元音，然后检查辅音。

Scanner scan = new Scanner(System.in);
System.out.print("Enter a sentence :");
String sentence = scan.nextLine();

String vowels = "aeiouAEIOU";
int vowelCount = 0;
int consCount = 0;
int i;
System.out.println("["+sentence+"]");
for (i = 0; i < sentence.length(); i += 1)
{
    char currentChar = sentence.charAt(i);
    boolean isVowel = false;
    int index;
    for (index = 0; index < vowels.length(); index += 1)
    {
        if (vowels.charAt(index) == (currentChar))
        {
            vowelCount++;
            isVowel =true;
            break;//we found the vowel, no need to keep looping
        }
    }

     if (!isVowel && Character.isLetter(currentChar))//vowel have not been found
    {
        consCount++;
    }

}

System.out.println("cons: " + consCount);
System.out.println("vowel: " + vowelCount);

Answer 6

使用此：

Scanner scan = new Scanner(System.in);
    System.out.print("Enter a sentence :");
    String sentence = scan.nextLine();

    String vowels = "aeiouAEIOU";
    int vowelCount = 0;
    int consCount = 0;
    int i;
    boolean flag;
      for(i = 0; i < sentence.length(); i += 1){
        char currentChar = sentence.charAt(i);
        flag=false;
        int index;
        for(index = 0; index < vowels.length(); index += 1){
          if(vowels.charAt(index) == (currentChar)){
            vowelCount++;
            flag=true;
            continue;
          }
        }
        if(!flag && Character.isLetter(currentChar) ){
            consCount++;
         }
      }
      System.out.println(consCount);
      System.out.println(vowelCount);

<强>更改

你在做什么：

 for(index = 0; index < vowels.length(); index += 1){
  if(vowels.charAt(index) == (currentChar)){
      vowelCount++;
    }else if(Character.isLetter(currentChar) && (vowels.charAt(index) == (currentChar))){
        consCount++;
  }

每当consCount与元音不匹配时，hello就会增加。例如，如果您尝试consCount，h仅会consCount++增加到10，因为它不会对任何元音进行数学运算。

所以我在这里做的是将flag放在第二个for循环之外并添加一个consCount++来检查它是否是元音。如果它是一个元音，那么{{1}}将被跳过，如果没有，那么它将被检查为后者被认为是一个辅音。

Answer 7

public static void main(String[] args) {

    Scanner stringin = new Scanner(System.in);
    String string1;
    System.out.println("Enter a string: ");
    string1 = stringin.nextLine();
    string1 = string1.toLowerCase();

    int count = 0;
    int vowels = 0;
    int consonants = 0;
    int spaces = 0;
    for (int i = 0; i < string1.length(); i++) {

        char letter = string1.charAt(i); 
        if (letter == 'a' || letter == 'e' || 
                letter == 'i' || letter == 'o' ||
                    letter == 'u')
        {
            vowels++;
        }
        else if  (letter == ' ')
        {
            spaces++;
        }
        else
        {
            consonants++;


        }
    }
    System.out.print("The number of vowels is " + vowels);
    System.out.print(" The number of consonants is " + consonants);
}

这是我的解决方案。我不得不对空间进行例外处理，因为它被视为辅音。目前正在上Java课程，所以我根本没有经验。

如何计算Java中String中辅音的数量？

7 个答案: