search string from bigger string

[Juhan Voolaid]

Hi
I have a big string. For example:
bigString="This is big string. Big string is not small";
and i have small string:
smallString="is";
and i have to find how many matches with small string i have in big one.
As a result there should be 2 matches. I have searched the whole string class doc and nothing simple has not come to my mind - it can't be hard.
So does anyone know any good methods to find a match of one string to another?
thnx.

[Igor Ko]

Is it your home work obviously ?
Something like this:

  int countSubstring(String str, String pattern){
    int counter = 0;
    int offset = 0;
    for(;   :wink:   { // Here should be empty for() - not Graemlins 
                 // Something does't work well in the forum !!!
      int ix = str.indexOf(pattern, offset);
      if( ix == -1 )
        break;
      offset = ix+1;
      counter++;
    }
    return counter;
  }

[ December 01, 2003: Message edited by: Igor Ko ]

[Juhan Voolaid]

Yes - it's my homework at school and i'm really out of time. I remember the teacher showd this in class and it was so obvious that i didn't write it down - now i'm in deep ...
I didn't understand the code quite clearly - so if u could add sine comments it would be nice.
Allso ... String.regionMatches() - i don't understand what are the parameters for this fx and can i use it?
ok thnx - need more help.

[Carlos Failde]

Take a look at these examples.

class StringMatches
{
  static String bigString="This is big string. Big string is not small";
  static String smallString="is";
  public static void main(String [] args)
  {
    int count = 0;
    for(int i=0; i < bigString.length() - smallString.length(); i++)
    {
      if( bigString.regionMatches(i, smallString, 0, smallString.length() ))
        count ++;
    }
    System.out.println(count);
  }
}

Output is 3. Matches all occurances of "is" even the one in "this"

class StringSplit
{
  static String bigString="This is big string. Big string is not small";
  static String smallString="is";
  public static void main(String [] args)
  {
    int count = 0;
    // split bigString into words by splitting about space character
    String [] split = bigString.split(" ");
    // check if a word matches smallString
    for(int i=0; i < split.length; i++)
    {
      if( split[i].equals(smallString) )
        count ++;
    }
    System.out.println(count);
  }
}

Output is 2.
But if bigstring had contained something like "Do not end a sentence with the word is." it would have returned 0 because of the fullstop.
By using regular expressions we can get around that:

import java.util.regex.Pattern;
import java.util.regex.Matcher;
class StringRegSplit
{
  static String bigString="This is big string. Big string is not small";
  static String smallString="is";
  public static void main(String [] args)
  {
    // Create a pattern to match smallString
    Pattern p = Pattern.compile(smallString);
    Matcher m = p.matcher(bigString);
    int count = 0;
    while( m.find() )
    {
      count ++;
    }
    System.out.println(count);
  }
}

Returns 3. Similar to the first attempt - trying to match "is" also matches "this"

import java.util.regex.Pattern;
class StringSpaceSplit
{
  static String bigString="This is big string. Big string is not small.";
  static String smallString="is";
  public static void main(String [] args)
  {
    // Create a pattern to match spaces, commas, and fullstops.
    Pattern p = Pattern.compile("[,.\\s]+");
    // Split bigString into words with the pattern
    String[] result = p.split(bigString);
    // find strings matching smallString
    int count = 0;
    for(int i =0; i < result.length; i++)
    {
      if( result[i].equals(smallString) )
        count++;
    }
    System.out.println( count );
  }
}

Returns 2

[Zeeman Chen]

I'm a newbie, I have a question I'd like to ask,
Is empty for loop " for(; a good programming style?

[fred rosenberger]

I would not use an empty for loop. I use a for loop when i know (or can calculate) the number of iterations i need.
if i want an infinate loop, with a break condition, i'd use

while(true)
   {
   //blah blah blah
   if (whatever)
        break;
   }

Better/worse/proper is a debate that will go on forever...

[John Guilliani]

Is there a way to count the number of times the smallString is found in reverse order as well? For example if the BigString = "Mississipi" and the SmallString = "is". The number of occurrences would be 4. Counting the occurrences of "si" as well.
M(is)(s(i)s)(si)pi.