Java - 计算字符串中的元音

Java - Counting the vowels in string

下面的代码应该根据以下原则计算字符串中的音节数:

  1. 应将单个元音或一组元音计为一个音节。
  2. 如果一个单独的 "e" 在字符串的末尾并且在字符串的其余部分有更多的元音,则 "e" 不是一个音节。
  3. 如果一个单独的"e"在末尾,并且"e"旁边有一个或多个元音,并且字符串的其余部分也有更多的元音,则"e"是一个音节。

我的代码执行前两条规则但不执行最后一条。有人可以帮我修改这段代码,使第三条规则也得到满足吗?

protected int countSyllables(String word) {
    String input = word.toLowerCase();
    int syl = 0;
    boolean  vowel  = false;
    int length = word.length();
    //check each word for vowels (don't count more than one vowel in a row)
    for(int i=0; i<length; i++) {
        if        (isVowel(input.charAt(i)) && (vowel==false)) {
            vowel = true;
            syl++;
        } else if (isVowel(input.charAt(i)) && (vowel==true)) {
            vowel = true;
        } else {
            vowel = false;
        }
    }
    char tempChar = input.charAt(input.length()-1);
    //check for 'e' at the end, as long as not a word w/ one syllable
    if ((tempChar == 'e')  && (syl != 1)) {
        syl--;
    }
    return syl;
}
protected int countSyllables(String word) {
    if(word.isEmpty()) return 0; //don't bother if String is empty

    word = word.toLowerCase();
    int      totalSyllables    = 0;
    boolean  previousIsVowel  = false;
    int      length = word.length();

    //check each word for vowels (don't count more than one vowel in a row)
    for(int i=0; i<length; i++) {
        //create temp variable for vowel
        boolean isVowel = isVowel(word.charAt(i));

        //use ternary operator as it is much simple (condition ? true : false)
        //only increments syllable if current char is vowel and previous is not
        totalSyllables += isVowel && !previousIsVowel ? 1 : 0;

        if(i == length - 1) { //if last index to allow for 'helloe' to equal 2 instead of 1
            if (word.charAt(length - 1) == 'e' && !previousIsVowel)
                totalSyllables--; //who cares if this is -1
        }

        //set previousVowel from temp 
        previousIsVowel = isVowel;
    }

    //always return 1 syllable
    return totalSyllables > 0 ? totalSyllables : 1;
}

测试 e 是否在结尾时,只需在删除音节之前添加一个条件。只需确保从末尾开始的一个字符不是元音即可。


解决方案

if ((tempChar == 'e')  && (syl != 1) && !isVowel(word.charAt(word.length()-2))) {
    syl--;
}

输出

public static void main(String[] args) {
    System.out.println(countSyllables("Canoe"));  // 2
    System.out.println(countSyllables("Bounce")); // 1
    System.out.println(countSyllables("Free"));   // 1
}