我正在研究Pig Latin项目，该项目需要更改用户输入的任何句子以便翻译成Pig Latin。我有转换，它的工作原理。但是我有标点问题。当我分割我的字符串以处理字符串中的每个单词时，标点符号会受到阻碍。我想知道一种方法，能够将字符串输入分成单个单词，但保留分隔符，然后能够正确地放回标点符号和空格？
谢谢
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Enter a word or phrase: ");
String convert = scanner.nextLine();
String punctuations = ".,?!;";
//convert = convert.replaceAll("\\p{Punct}+", ""); //idk if this is useful for me
String finalSentence = "";
if (convert.contains(" ")) {
String[] arr = convert.split("[ ,?!;:.]+");
for (int index = 0; index < arr.length; index++) {
if (vowel(arr[index]) == true) {
System.out.println(arr[index] + "yay");
finalSentence = (finalSentence + arr[index] + "yay ");
} else {
System.out.println(newConvert(arr[index]));
finalSentence = (finalSentence + newConvert(arr[index]) + " ");
}
}
答案 0 :(得分：0)
...
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使用 public static void main(String[] args) {
String convert = "The quick? brown!!fox jumps__over the lazy333 dog.";
StringBuilder finalSentence = new StringBuilder();
List<String> tokens = Arrays.asList(convert.split(""));
Iterator<String> it = tokens.iterator();
while (it.hasNext()) {
String token = it.next();
StringBuilder sb = new StringBuilder();
while (token.matches("[A-Za-z]")) {
sb.append(token);
if (it.hasNext()) {
token = it.next();
} else {
token = "";
break;
}
}
String word = sb.toString();
if (!word.isEmpty()) {
finalSentence.append(magic(word));
}
finalSentence.append(token);
}
//prints "The1 quick1? brown1!!fox1 jumps1__over1 the1 lazy1333 dog1."
System.out.println(finalSentence.toString());
}
private static String magic(String word) {
return word + 1;
}
方法进行Pig拉丁语翻译。
答案 1 :(得分：0)
我为Pig Latin翻译定义了两种方法： convert_word_to_pig_latin 方法是将每个单词转换为Pig Latin， convert_sentence_to_pig_latin 方法用于使用convert_word_to_pig_latin的句子方法
def convert_word_to_pig_latin(word)
vowels = "aeiou"
punctuations = ".,?!'\":;-"
if vowels.include?(word[0])
return word
else
if punctuations.include?(word[-1])
punctuation = word[-1]
word = word.chop
end
first_vowel_index = word.chars.find_index { |letter| vowels.include?(letter) }
new_word = word[first_vowel_index..-1] + word[0...first_vowel_index] + "ay"
return punctuation ? new_word += punctuation : new_word
end
end
def convert_sentence_to_pig_latin(sentence)
sentence_array = sentence.split(" ")
sentence_array.map { |word| convert_word_to_pig_latin(word) }.join(" ")
end
注意：请随意添加任何其他标点符号。
最后，这是我的RSpec，以确保我的两个方法都通过所有测试：
require_relative('../pig_latin')
describe 'Converting single words to Pig Latin' do
word1 = "beautiful"
word2 = "easy"
word3 = "straight"
it "converts word to Pig Latin" do
expect(convert_word_to_pig_latin(word1)).to eq "eautifulbay"
end
it "does not change word if it begins with a vowel" do
expect(convert_word_to_pig_latin(word2)).to eq "easy"
end
it "converts word to Pig Latin" do
expect(convert_word_to_pig_latin(word3)).to eq "aightstray"
end
end
describe 'Converting a sentence to Pig Latin' do
sentence1 = "Make your life a masterpiece; imagine no limitations on what you can be, have, or do."
sentence2 = "The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty."
it "converts motivational quote from Brian Tracy to Pig Latin" do
expect(convert_sentence_to_pig_latin(sentence1)).to eq "akeMay ouryay ifelay a asterpiecemay; imagine onay imitationslay on atwhay ouyay ancay ebay, avehay, or oday."
end
it "converts motivational quote from Winston Churchill to Pig Latin" do
expect(convert_sentence_to_pig_latin(sentence2)).to eq "eThay essimistpay eessay ifficultyday in every opportunity. eThay optimist eessay ethay opportunity in every ifficultyday."
end
end