Thread: Recursive Sentence Finder

Hi all, I have a programming assignment that is described as follows:

Use recursion to implement a method [B]boolean find(String t)[/B] that tests whether a string is contained in a sentence:
[B]Sentence s = new Sentence("Mississippi!");
boolean b = s.find("sip"); // Returns true[/B]

Here is what I have done so far in my Sentence class:

import java.util.ArrayList;
 
public class Sentence
{
	public Sentence( String newWord )
	{
		word = newWord;
	}
 
	public boolean find( String find )
	{	
		if( !(word.substring( 0, find.length() ).equalsIgnoreCase(find) ) )
		{
			String shorterWord = word.substring( 1, word.length() );
			Sentence shorterSentence = new Sentence( shorterWord );
			shorterSentence.find(find);
		}
		else if( word.substring( 0, find.length() ).equalsIgnoreCase(find) )
		{
			result = true;
		}
		else if( word.length() == find.length() && !(word.substring( 0, find.length() ).equalsIgnoreCase(find) ) )
		{
			result = false;
		}
 
		return result;
	}
 
	private boolean result;
	private String word;
}

What I have done is I have tested the substring from the start of the word to the length of word I am trying to find. If it does not match, then I take away the first letter to form a shorterWord, create a new setence and use the find() method on this word.

I'm pretty new to recursive thinking and I thought I had done this correctly, but it does not yield the results I'm looking for. Can anyone help me with this? Thanks in advance.

Here is my tester class:

public class SentenceFinder
{
	public static void main(String[] args)
	{
		Sentence s = new Sentence( "Mississippi!" );
		boolean b = s.find( "sip" );
		System.out.println( b );
	}
}

Thank you! It's been bothering me for a while now. Would you mind showing me how you would tidy up my code? I'd like to get into the right habits ahead of time so that I can continue practicing good programming.

Hi, thanks for the schooling. I did realize that the algorithm was a bit off and revised my code, I just didn't post the revised code:

import java.util.ArrayList;
 
public class Sentence
{
	public Sentence( String newWord )
	{
		word = newWord;
	}
 
	public boolean find( String find )
	{	
		if( word.length() == find.length() && !(word.substring( 0, find.length() ).equalsIgnoreCase(find) ) )
		{
			result = false;
		}
		else if( !(word.substring( 0, find.length() ).equalsIgnoreCase(find) ) )
		{
			String shorterWord = word.substring( 1, word.length() );
			Sentence shorterSentence = new Sentence( shorterWord );
			result = shorterSentence.find(find);
		}
		else if( word.substring( 0, find.length() ).equalsIgnoreCase(find) )
		{
			result = true;
		} 
 
		return result;
	}
 
	private boolean result;
	private String word;
}

I still think that the third else if cause (the first one in the revised code above) is necessary. For example, if the word was "robot" and we were searching for "bot":

(robot, bot) = (obot, bot)
(obot, bot) = (bot, bot)

If I took out that else if clause, word.length() would surely equal find.length() but it wouldn't check to see whether the words matched and would just denote to false. Adding !(word.substring(0, find.length() ).equalsIgnoreCase(find) ) ) allows it to check for equality (or non-equality in this case). That's the only way I could think of to get it to skip over this step if the words matched and go into the last else if clause (the one that sets result to true).

Is my logic off here?