让计数器只计算字母，而不是空格

Question

看起来很简单，但我是编码新手。我正在尝试设置一个计数器来计算用户输入字符串的每个字母，但我不想计算空格。

我看到很多人推荐使用 replaceAll() 函数，但我还没有了解它，也不知道如何/在哪里定义它。想知道是否有另一种方法，如果没有，如何调用/replaceAll(" ", "") 的语法，以及它在代码中的位置。

import java.util.Scanner;

public class Vowel_Counter
{

    public static void main (String[] args)
    {
        int aCount= 0 , eCount= 0, iCount= 0;
        int total= 0, totalConsonant= 0, count = 0;
        char vowels= 0, i = 0;


        String userInput;

        System.out.println("Please enter a string of any lenth, or character combination");
        Scanner scan = new Scanner (System.in);

    String userIntput = Vowel_Counter.replace(" ","");

        userInput = scan.nextLine(); 



        for (count = 0; count < userInput.length(); count++) 
        {


            vowels = userInput.charAt(count); 

            switch (vowels)
            { 
            case 'a':
                aCount++;
                break;
            case 'e':
                eCount++;
                break;
            case 'i':
                iCount++;
                break;

            }

        }
            System.out.println("There are " +aCount+ " a's in that string"); 
            System.out.println("There are " +eCount+ " e's in that string");
            System.out.println("There are " +iCount+ " i's in that string");

           total = userInput.length();


            System.out.println("The total lenth is " + total + " charactors");

        totalConsonant = total - aCount -eCount -iCount;

System.out.println("There are " + totalConsonant+ " non-vowels in that string");
}

}

如果输入的字符串是“我爱狗”，它应该输出 “有 0 个” “有 1 个 e” “有 1 个我” “总长度为 9 个字符” “有 5 个非元音”

（代码确实包含每个元音的计数器，但我已经为这个问题缩短了它）

Answer 1

在我看来，最简单（最容易理解）的方法是检查字符是否为空格，如果是则跳过：

int charCount = 0;
for (count = 0; count < userInput.length(); count++) 
    {
        input = userInput.charAt(count); 

        if(input == ' ')
            continue;

        charCount++;

        switch (input)
        { 
        case 'a':
            aCount++;
            break;
        case 'e':
            eCount++;
            break;
        case 'i':
            iCount++;
            break;
        //(...other two cases)
        default:
            totalConsonant++;
        }
    }

System.out.println("There are " +aCount+ " a's in that string"); 
System.out.println("There are " +eCount+ " e's in that string");
System.out.println("There are " +iCount+ " i's in that string");
// ... other two cases


System.out.println("The total length is " + charCount + " characters");

System.out.println("There are " + totalConsonant + " non-vowels in that string");

我可能还建议使用 replaceAll()，但如果您要求不同的方式，这应该以一种非常简单的方式进行。 'continue' 关键字将跳过循环的其余部分并继续下一次迭代。

我也只是在默认情况下为 TotalConsonant 添加了一个计数，所以你不必在最后做凌乱的减法。

同样，当然不是最有效的，但它显示了一种非常简单的方法。

使用replaceAll()：

userInput = scan.nextLine().replaceAll(" ", "");

for (count = 0; count < userInput.length(); count++) 
{
    vowels = userInput.charAt(count); 

    switch (vowels)
    { 
        case 'a':
            aCount++;
            break;
        case 'e':
            eCount++;
            break;
        case 'i':
            iCount++;
            break;
        // ...
        default:
            totalConsonant++;
    }
}

System.out.println("There are " +aCount+ " a's in that string"); 
System.out.println("There are " +eCount+ " e's in that string");
System.out.println("There are " +iCount+ " i's in that string");
// ... other two cases

System.out.println("The total length is " + userInput.length() + " characters");

System.out.println("There are " + totalConsonant + " non-vowels in that string");