java - 如何检查数组是否包含特定数量的值？

Question

import java.util.Scanner;

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

获取数组的大小:

        System.out.println("Type how many words will be typed: ");
        int input = scan.nextInt();    

用字符串值填充数组

        String[] ar1 = new String[input];
        for(int i = 0; i < ar1.length; i++){
            System.out.println("Type the elements of array with words: ");
            ar1[i] = scan.next();      
        }

程序输出:

        System.out.println( input + " words are typed and " + 
        countVowels(ar1) + 
         " of them contain more than 3 vowels.");
    }

计算元音的方法:

    public static int countVowels(String[] ar1){  // this method counts 

        int a = 0;
        String[] ar2 = new String[]{"a", "e", "i", "u", "y", "o"};
        for(int i = 0; i < ar1.length; i++){
            for(String s : ar2){
                if(ar1[i].toLowerCase().contains(s)){
                    a++;
                }
            }
        }
        return a;
    }
}

上面的方法是检查元音，但我不知道如何让它检查是否元音超过3个。

Answer 1

另一种使用 replaceAll 方法的解决方案。主要思想是从 word.length() 中减去没有元音的相同字长。并检查差异。

public static int countVowels(String[] ar1){
    int a = 0;
    for (String word : ar1) {
        int i = word.length() - word.toLowerCase().replaceAll("[aeyiuo]", "").length();
        if (i >= 3) {
            a++;
        }
    }
    return a;
}

或者您可以按照@pkgajulapalli 的建议使用matches()。用stream api可以很简洁:

long count = Arrays.stream(words)
        .filter(s -> s.toLowerCase().matches("(.*[aeyiuo].*){3,}"))
        .count();