Find word in User Content without using .split() or StringTokenizer

Question

I'm working on a program that ask the user to input a phrase and an integer. The integer is used to identify which word will be return from the phrase. For example, if they enter 5, the program should return to the user the fifth word in the sentence.

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

This is my work so far, there is a main to output,

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;

And I was thinking a lot on how to work on the middle part and my thought are if the user pick numWord = 5, then in order to return the fifth word in that sentence, I'm gonna need to use "newPhrase.substring(space 4th +1, space 5th)". And this is where I stuck because I don't know how to start, and how to get to space 4th.

look into `String.split()`... it will makes this a bunch easier — RobOhRob, Oct 25 '19 at 17:32
You state in the title that you do not want to use split, but you have not given a reason why. — fishinear, Oct 25 '19 at 17:52
Hi, I'm sorry that I forgot to state the reason for it. I don't want to use split and StringTokenizer because first I haven't learnt about them and I know that it is gonna way easy if I use them. But at my level right now at a beginner, I want to figure out if there is any ways that just using substring and loop can make this works. — evane, Oct 25 '19 at 18:16
Hm, a valid sentence would be e.g. `No, it was 'no?', and not 'yes!' - ok?`, now the result would be `No`, `it` `was` `'no?,'` etc. ,you'd have to treat `' ,?!-` also as whitespace, wouldn't you, and cover multiple blanks in a row? You could use the regex-matcher approach from [here](https://stackoverflow.com/questions/6020384/create-array-of-regex-matches) with this regex to get single words `"[A-Za-z0-9]*"`. Still you'd have issues with words like `you'd`, that's one word, right? — Curiosa Globunznik, Oct 25 '19 at 18:42

Abhishek pruthvi V M · Answer 1 · 2019-10-25T18:27:30.153

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
  }

This code should work for you. If that's the case accept the answer.

Curiosa Globunznik · Answer 2 · 2019-10-26T07:20:00.267

If you want to go really low level, then you can go lower than subString and operate on single characters. This way its easy to skip other characters than blank. Its also a step towards the way regular expression get executed by transforming them to finite state automatons.

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;
}

Find word in User Content without using .split() or StringTokenizer

2 Answers2