Java 如何在字符串中找到元音，并在屏幕上打印元音最多的单词？

How do I find vowels in a string, and print the word with the most vowels on the screen?

提问by user2993612

I need to find all vowels in a user inputted string, and then print the word with the most vowels on the screen.
The program uses user input.
The user types in a string of words in lowercase.
e.g.

我需要在用户输入的字符串中找到所有元音，然后在屏幕上打印元音最多的单词。
该程序使用用户输入。
用户输入一串小写的单词。
例如

"I like java programming"

“我喜欢java编程”

and it should read out:

它应该读出：

programming

编程

I tried splitting the string into different words, and that works.
I only don't know how to apply a "for" loop to search in the different words. I need to work in methods, so this is the method I used to find vowels in the string:

我尝试将字符串拆分为不同的单词，效果很好。
我只是不知道如何应用“for”循环来搜索不同的单词。我需要在方法中工作，所以这是我用来在字符串中查找元音的方法：

public void findvowel(){
    for(int index = 0;index < word.length();index++){
    char vowel = word.charAt(index);
    if( (vowel == 'a')||
        (vowel == 'e')||
        (vowel == 'i')||
        (vowel == 'o')||
        (vowel == 'u')){
            System.out.println(vowel);
            }
        }
    }

But I know this doesn't work. Can you people help me?

但我知道这行不通。你们能帮我吗？

回答by Elliott Frisch

This should be a good starting point; note that the method name now really says what it does -

这应该是一个很好的起点；请注意，方法名称现在真正说明了它的作用 -

// public static int findvowel(String word) {
public static int getVowelCount(String word) {
  int count = 0;
  if (word != null) {
    word = word.trim().toLowerCase();
  }
  if (word == null || word.length() < 1) {
    return count;
  }
  for (int index = 0; index < word.length(); index++) {
    // That Fred he's a
    char fred = word.charAt(index);
    if ((fred == 'a') || (fred == 'e')
        || (fred == 'i') || (fred == 'o')
        || (fred == 'u')) {
      ++count;
      }
  }
  System.out.println("For the word \"" + word
      + "\" there are " + count + " vowels");
  return count;
}

回答by Little Child

public class MaxVowels {
    public static void main(String[] args) {
        String sentence = "This is a loooooooooong sentence";
        int maxVowelCount = 0;
        String wordsWithMostVowels = null;
        String[] words = sentence.split(" ");

        for(String word : words){
            int vowelCount = 0;
            word = word.toLowerCase();
            for(int i = 0; i < word.length() ; i++){
                char x = word.charAt(i);
                if(x == 'a' || x == 'e' || x == 'i' ||
                   x == 'o' || x == 'u'){
                    vowelCount++;
                }
            }
            if(vowelCount > maxVowelCount){
                maxVowelCount = vowelCount;
                wordsWithMostVowels = word;
            }
        }
        System.out.println("Word with most vowels is: " + wordsWithMostVowels);
    }
}  

The code is fairly straightforward and needs no explanation =)
The code ignores the case where two words have the same number of vowels. In this case, the first word will be used as the word with most vowels.

代码相当简单，无需解释 =)
代码忽略了两个单词具有相同元音数量的情况。在这种情况下，第一个单词将用作元音最多的单词。

回答by hthserhs

You're going in the right direction. Few things:

你正朝着正确的方向前进。几样东西：

You don't need to print vowels. You will be counting number of vowels in all the words. As you will be doing it one word at a time, you want to remember counts for words that came earlier. A better strategy is to remember only the word having maximum number of vowels. Whenever you find a word with more vowels, you update your result.

您不需要打印元音。您将计算所有单词中元音的数量。由于您将一次做一个单词，因此您想记住之前出现的单词的计数。更好的策略是只记住具有最大元音数的单词。每当你发现一个带有更多元音的单词时，你就更新你的结果。

You can use fieldsto remember the word with maximum number of vowels along with the number:

您可以使用字段来记住具有最大元音数的单词以及数字：

String wordWithMaxVowels;
int maxNumberOfVowels;

Suppose at this instance you're working on a single word. You need a localvariable to keep a count of vowels in this word.

假设在这种情况下您正在处理单个word. 您需要一个局部变量来保持 this 中元音的数量word。

int vowelCount = 0;
// Your code to loop over the word but remember
// to increase vowelCount if you find a vowel:
// vowelCount++;

Finally the check to see if this number is greater than the maximum we have so far. Update the fields if that's the case:

最后检查这个数字是否大于我们目前拥有的最大值。如果是这种情况，请更新字段：

if(vowelCount > maxNumberOfVowels) {
    wordWithMaxVowels = word;
    maxNumberOfVowels = vowelCount;
}

Another tip follows. To check if a character cis a vowel, you can:

另一个提示如下。要检查字符c是否为元音，您可以：

if ("aeiouAEIOU".indexOf(c) != -1) {
    vowelCount++;
}

回答by frankie liuzzi

Your findVowel() method is almost there. Why are you outputting the vowels when you're supposed to count them instead? Instead of findVowel(), i think you want something called countVowels() that returns the amount of vowels in a word. Something like this:

您的 findVowel() 方法几乎就在那里。当你应该计算它们时，为什么要输出元音？而不是 findVowel()，我认为你想要一个叫做 countVowels() 的东西，它返回一个单词中元音的数量。像这样的东西：

    public int countVowels(String word){
      int count = 0;
      for(int index = 0;index < word.length();index++){
        char vowel = word.charAt(index);
        if( (vowel == 'a')||
                (vowel == 'e')||
                (vowel == 'i')||
                (vowel == 'o')||
                (vowel == 'u')){
            count++;
        }
      }
      return count;
    }

this way, you can call countVowels() on every word in the sentence, and keep track of the word with the most vowels so far. Ex:

这样，您可以对句子中的每个单词调用 countVowels()，并跟踪到目前为止元音最多的单词。前任：

String sentence = "This is a sentence.";
String[] words = sentence.split(" ");  //splits sentence into string array by spaces

String maxStringSoFar = "";
int maxStringVowelCount = 0;
for(String s : words)
{
     int currentWordVowelCount = countVowels(s);
     if(currentWordVowelCount > maxStringVowelCount)
     {
          maxStringSoFar = s;
          maxStringVowelCount = currentWordVowelCount;
     }
}

回答by Nitika Goswami

Hey I have some Different answer-----

嘿，我有一些不同的答案-----

class strDemo2
{
    public static void main(String args[])
    {
    String s1=new String("The Ghost of The Arabean Sea");


        char c1[]=new char[30];
        char c2[]={'a','e','i','o','u'}; 
         s1.getChars(0,28,c1,0);
         int pf=0,pl=0;
        for(int i=0;i<s1.length();i++)
        {
           for(int j=0;j<5;j++) 
           {
              if(c1[i]==c2[j])   
              {
          System.out.println("vowel found at:-> "+(i+1)+" which is "+c2[j]);
              }    
           }  
        }               

   }
}

回答by Harsh Raj

public class Test2{
public static void main(String[] args) {
    String message = "words containig mooooost vowels";
    String wordWithMostVowel = null;
    int maxVowelCount = 0;
    String tests[] = message.split(" ");
    int totalVowel = 0;
    for(String test : tests){
        int vowelCount = 0;
        test = test.toLowerCase();
        for(int i=0;i<test.length();i++){
            switch(test.charAt(i)){
            case 'a':
            case 'e':
            case 'i':
            case 'o':
            case 'u':
            vowelCount++;
            }
            if(vowelCount > maxVowelCount){
                maxVowelCount = vowelCount;
                wordWithMostVowel = test;
            }
        }
        totalVowel = totalVowel+vowelCount;
    }
    System.out.println("total vowels "+totalVowel+" word with max vowel is "+wordWithMostVowel);
}

}

}

回答by sivananda

public static void Vowels(){
        String Name = "seleniumtesting";
        String Vowels = "aeiou";
        char[] N = Name.toCharArray();
        for(int i=0; i<N.length; i++){

            if(Vowels.contains(String.valueOf(N[i]))){
                System.out.println("It is a Vowel : "+ N[i]);

            }
        }
    }