c语言拉丁猪的代码,抛光我的猪拉丁语翻译器（代码）（Java）

港控/mmm° 2022-10-14 14:52 196阅读 0赞

我的任务是使用递归创建猪拉丁语翻译器，可以翻译句子。规则如下：抛光我的猪拉丁语翻译器(代码)(Java)

如果英语中没有元音字母，那么pigLatinWord只是英文字+“ay”。 (有10个元音：'a'，'e'，'i'，'o'和'u'，以及它们的大写字母。'y'不被认为是用于这个任务的元音，即我变成了myay，为什么变成了whyay等)

否则，如果englishWord以元音开头，那么pigLatinWord只是englishWord +“yay”。

否则(如果englishWord中有一个元音，但不与元音开始)，然后pigLatinWord是端+开始+“AY”，其中结束和开始的定义如下： -

让我们开始使用所有的englishWord(但不包括)其第一个元音。 让我们结束从其第一个元音的所有英语词汇。 但是，如果英语字词大写，那么大写结尾和“非大写字母”开始。

这里是我到目前为止的代码(抱歉怪异的格式，我的意见弄乱)：

/\*Recursively Translate a String (without punctuation or numerical characters) to Pig Latin\*/

//prep the string for translation and submit it to be translated

public static String translate(String finished) \{

finished.trim(); //Trim the String of whitespace at the front and end

finished += " "; //Because of the recursive method I use, the string must have a

//space at the end

finished = translateSentence(finished); //recursively translate the string

finished.trim(); //trim the whitespace added earlier

return finished; //Return the string

//recursively submits each word in the string to the translator, then

//returns the translated sentence

private static String translateSentence(String finished) \{

if (finished.length() == 0) \{ //the base condition is met when each word in the string

return finished; //has been sent to the translator (string is empty)

else \{

return (translateWord(finished.substring(0, finished.indexOf(' '))) + " "

\+ translateSentence(finished.substring(finished.indexOf(' ') + 1)));

/\*If the base condition is not met, the method returns the first word of the string

\* (translated) and a space, then submits the rest of the

\* string back to the method. The first word is defined as the beginning

\* of the string up until the first space. The rest of the string

\* starts one character after the space. \*/

//Checks the submitted word for vowels and vowel placement, and translates accordingly

private static String translateWord(String stringA) \{

if (stringA.indexOf('a') == -1

&& stringA.indexOf('e') == -1 //Checks for presence of any vowels

&& stringA.indexOf('i') == -1 //if no vowels are found

&& stringA.indexOf('o') == -1 //the word + ay is returned

&& stringA.indexOf('u') == -1

&& stringA.indexOf('A') == -1

&& stringA.indexOf('E') == -1

&& stringA.indexOf('I') == -1

&& stringA.indexOf('O') == -1

&& stringA.indexOf('U') == -1) \{

return stringA + "ay";

if (stringA.charAt(0) == 'a'

|| stringA.charAt(0) == 'e' //checks if there is a vowel at the start

|| stringA.charAt(0) == 'i'//of the string. if there is a vowel

|| stringA.charAt(0) == 'o' //it returns the word + yay

|| stringA.charAt(0) == 'u'

|| stringA.charAt(0) == 'A'

|| stringA.charAt(0) == 'E'

|| stringA.charAt(0) == 'I'

|| stringA.charAt(0) == 'O'

|| stringA.charAt(0) == 'U') \{

return stringA + "yay";

/\* if the word has a vowel that isn't at the start, the part of the string

\* before the first vowel is moved to the end of the vowel, and "ay" is added.

\* However, if the first character in the word is capitalized, the first vowel becomes

\* uppercase and the former first character in the word becomes lowercase \*/

else \{

if (Character.isUpperCase(stringA.charAt(0))) \{

return Character.toUpperCase(stringA.charAt(firstVowel(stringA, 0)))

\+ stringA.substring(firstVowel(stringA, 0) + 1, stringA.length())

\+ Character.toLowerCase(stringA.charAt(0))

\+ stringA.substring(1, firstVowel(stringA, 0)) + "ay";

else \{

return stringA.substring(firstVowel(stringA, 0), stringA.length())

\+ stringA.substring(0, firstVowel(stringA, 0)) + "ay";

//Recursively determines the index number of the first vowel in a given word

//0 must always be submitted as int x

public static int firstVowel(String stringA, int x) \{

if (x > stringA.length() - 1) \{ //if the index number becomes greater than the length

return -1; //of the string, -1 (no vowels) is returned

if (stringA.charAt(x) == 'a'

|| stringA.charAt(x) == 'e' //the base condition is met when the character

|| stringA.charAt(x) == 'i' //at the current index number is a vowel

|| stringA.charAt(x) == 'o' //and the index number is returned

|| stringA.charAt(x) == 'u'

|| stringA.charAt(x) == 'A'

|| stringA.charAt(x) == 'E'

|| stringA.charAt(x) == 'I'

|| stringA.charAt(x) == 'O'

|| stringA.charAt(x) == 'U') \{

return x;

else \{

return firstVowel(stringA, x + 1); //otherwise, the string and the index number

\} // + 1 are submitted back to the method

这给了我所需要的输出(“为什么你好”变成了“Whyay ellohay erethay”) 但现在它不能处理标点符号。基本上我正在寻找的是任何提示或帮助让我的代码处理标点符号，或任何方式来改善我的代码(仍然使用递归)。

2012-11-22

Philip

\+1

一个改进是将所有元音放在一个字符串常量中，并使用'contains()'检查一个字符是否为元音。此外，'firstVowel(stringA，0)'可以被提取到一个局部变量中，这将进一步提高可读性。 –

\+0

你能澄清有关contains()方法的部分吗？我不太了解如何将字符串常量，contains()和我正在检查的字符串关联起来。谢谢。 –

\+1

我的意思是这样的：'private static final String VOWELS =“aeiouAEIOU”;'，然后当你检查元音时，你做'private static boolean isVowel(char ch)\{return VOWELS.contains(String.valueOf (ch)); \}'。阅读起来比“if”陈述容易得多。 –