Java pig拉丁语类节目-元音问题

Java pig拉丁语类节目-元音问题,java,Java,我正在编写一个有两条规则的程序: 1.如果单词的第一个字符是元音,则将其移至单词末尾。 2.如果单词的第一个字符是辅音,则将其移动到单词的末尾并附加“ae” import java.util.Scanner; public class Program5 { public static void main(String[] args) { // TODO Auto-generated method stub Scanner scanner = new S

我正在编写一个有两条规则的程序:
1.如果单词的第一个字符是元音,则将其移至单词末尾。
2.如果单词的第一个字符是辅音,则将其移动到单词的末尾并附加“ae”

import java.util.Scanner;

public class Program5 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner scanner = new Scanner(System.in);
        System.out.print("Please enter a sentence: ");
        String english = scanner.nextLine();
        String piggy = piggyEnglish(english);
        System.out.print("Translated: " + piggy);
    }
     private static String piggyEnglish(String s) {
            String piggy = "";
            int i = 0;
            while (i<s.length()) {

              while (i<s.length() && !isLetter(s.charAt(i))) {
                piggy = piggy + s.charAt(i);
                i++;
              }

              if (i>=s.length()) break;

              int begin = i;
              while (i<s.length() && isLetter(s.charAt(i))) {
                i++;
              }

              int end = i;
              piggy = piggy + piggyWord(s.substring(begin, end));
            }
            return piggy;
          }

     private static boolean beginsWithVowel(String word){
              String vowels = "aeiou";
              char letter = word.charAt(0);
              return (vowels.indexOf(letter) != -1);
              }

     private static boolean isLetter(char c) {
        return ( (c >='A' && c <='Z') || (c >='a' && c <='z') );
      }

    private static String piggyWord(String word) {
            int split = firstVowel(word);
            if(beginsWithVowel(word)) {
                return word.substring(split) + word.substring(0, split);
            } else {
            return word.substring(split) + word.substring(0, split)+"ae";
            }
    }       

    private static int firstVowel(String word) {
        word = word.toLowerCase();
        for (int i=0; i<word.length(); i++)
          if (word.charAt(i)=='a' || word.charAt(i)=='e' ||
              word.charAt(i)=='i' || word.charAt(i)=='o' ||
              word.charAt(i)=='u')
            return i;
        return 0;
    }
}
然而,我得到的是:

Please enter a sentence: today is a beautiful day
Translated: odaytae is a eautifulbae aydae

基本上,它不会翻译任何以元音开头的单词。我认为问题源于piggyWord方法,但我不确定。我可以得到关于如何修复此问题的任何提示吗?

错误在于
piggyWord
函数:

private static String piggyWord(String word) {
        int split = firstVowel(word);
        if(beginsWithVowel(word)) {
            return word.substring(split + 1) + word.substring(0, split + 1); //Since vowel is in 1st place, substring(0,0) returns empty string.
        } else {
        return word.substring(split) + word.substring(0, split)+"ae";
        }
}   

根据您的规则,您不需要方法firstVowel()来获取单词中第一个誓言的索引,因为您只需要知道单词中的第一个字符是否为誓言

因此,只需将piggyWord方法更改为以下方法即可解决您的问题:

private static String piggyWord(String word) {
    if(beginsWithVowel(word)) {
        return word.substring(1) + word.substring(0, 1);
    } else {
        return word.substring(1) + word.substring(0, 1)+"ae";
    }
}
或者更简单地说:

private static String piggyWord(String word) {
    String result = word.substring(1) + word.substring(0, 1);
    return beginsWithVowel(word) ? result : result + "ae";
}

因为您总是必须将单词的第一个字符移到末尾,唯一的问题是您是否需要在末尾附加一个额外的“ae”。

如果只涉及第一个字母,而不是第一个元音,那么如果元音位于第一个位置,您可以返回1

private static int firstVowel(String word) {
    word = word.toLowerCase();
    for (int i=0; i<word.length(); i++)
      if (word.charAt(i)=='a' || word.charAt(i)=='e' ||
          word.charAt(i)=='i' || word.charAt(i)=='o' ||
          word.charAt(i)=='u')
        return 1;
    return 0;
}
private static int first元音(字符串字){
word=word.toLowerCase();
对于(int i=0;i
private static int firstVowel(String word) {
    word = word.toLowerCase();
    for (int i=0; i<word.length(); i++)
      if (word.charAt(i)=='a' || word.charAt(i)=='e' ||
          word.charAt(i)=='i' || word.charAt(i)=='o' ||
          word.charAt(i)=='u')
        return 1;
    return 0;
}