在不使用.split（）或StringTokenizer的情况下查找用户内容中的单词

Question

我正在开发一个程序，要求用户输入短语和整数。整数用于标识从短语返回哪个单词。例如，如果他们输入5，则程序应将句子中的第五个单词返回给用户。

System.out.println("Your word is: " +combineString(phrase,numWord));

到目前为止，这是我的工作，有一个主输出，

public static String combineString(String newPhrase, int newNum) {
  int countWords = 0;
  String word = "";

  //words count. I'll +1 everytime using countWord the match the amount of words
  for(int i=0; i< newPhrase.length(); i++) {
     if(newPhrase.charAt(i) == ' ') {
        countWords++;             
     }
  }  

  //return the last word. Ex: 15 words in a phrase if user pick the 18th word it will return the 15th word.
  if(countWords+1 < newNum  || countWords+1 <= newNum) {
     word += newPhrase.substring(newPhrase.lastIndexOf(' ')+1, newPhrase.length()-1);
  }
  else if(newNum <=0) { //return null if the user pick 0 or less than 0
     word += null;   
  }           
  return word;

并且我在中间部分进行了很多思考，我的想法是，如果用户选择numWord = 5，那么为了返回该句子中的第五个单词，我需要使用“ newPhrase。 substring（space 4th +1，space 5th）”。这就是我坚持的地方，因为我不知道如何开始，也不知道如何到达太空4号。

Answer 1

public static String combineString(String newPhrase, int newNum) {
     if(newNum<=0)
        return null;
     String word = "";
     String [] match = new String[newNum];

    int j =0;
    for(int i=0; i< newPhrase.length(); i++) {
        word = word + newPhrase.charAt(i);
        if(newPhrase.charAt(i) == ' ') { 
           match[j] = word;
           if(j+1 == newNum) {
              return word; // returns the specified word
           } 
           j++;
           word = "";    
       }
    } 
    return word; //returns last word
  }

此代码应为您工作。如果是这样，请接受答案。

Answer 2

如果想降到很低的水平，则可以降到subString以下并使用单个字符进行操作。这样，很容易跳过空格以外的其他字符。通过将正则表达式转换为有限状态自动机，这也朝着执行正则表达式的方式迈出了一步。

enum ScanState {WHITESPACE, WORD}

private final static Set<Character> whitespace = new HashSet<>(Arrays.asList('"', ',', '.', '?', '!', '-', ';', ' '));

@Test
public void testTokenize() {
    char[] text = "No, it's been \"yes?\", and not \"no!\" - hasn't it?".toCharArray();
    List<String> expected = Arrays.asList("No", "it's", "been", "yes", "and", "not", "no", "hasn't", "it");
    assertEquals(expected, tokenize(text));
}

private List<String> tokenize(char[] text) {
    List<String> result = new ArrayList<String>();
    char[] word = new char[256];
    int maxLetter = 0;
    ScanState prevState = ScanState.WHITESPACE;

    for (char currentChar : text) {
        ScanState currState = whitespace.contains(currentChar) ? ScanState.WHITESPACE : ScanState.WORD;

        if (prevState == ScanState.WORD && currState == ScanState.WORD) {
            word[maxLetter++] = currentChar;
        }
        if (prevState == ScanState.WORD && currState == ScanState.WHITESPACE) {
            word[maxLetter++] = currentChar;
            result.add(String.valueOf(word, 0, maxLetter - 1));
        }
        if (prevState == ScanState.WHITESPACE && currState == ScanState.WORD) {
            maxLetter = 0;
            word[maxLetter++] = currentChar;
        }
        prevState = currState;
    }
    return result;
}