The cleanest way I can think of to recognize character clusters like those is to separate the string into an array of smaller strings (instead of a char[] array).  Most would be one character, but some of them two or three.

In other words, if your maximum char cluster length is 3 chars, check the first 3 chars of your string against all the 3-char clusters, then check the first 2 chars of the string against all the 2-char clusters, and if nothing is found, just save the first char.  Then remove however many were matched from the string, and repeat.

Thanks for replying!

I've been pretty much research on the sounds of the Dutch language. I have a database containing all sounds which occur in the Dutch language (about 11 pages of words). It looks like this:
M      mbt            ambt
     mbts      ambst
     md            beemd, hemd
     mds            hemds

vowels:
e                  bode, bewind.
e                  stevig, hevig.
e                  steel, been.
i                  in, zing.
y                  byssus.
i                  tiran.
                             
y                  Myronus.
iaau                  miaauwen

With this i can split the words into parts.

this is what i got right now:

Code:

import java.util.ArrayList;
import processing.video.*;
Movie myMovie;

ArrayList spc = new ArrayList ();
ArrayList WDstA1 = new ArrayList ();
ArrayList WDnrA1 = new ArrayList ();
String WDst1 = "";
String WDnr1 = "";
int i = 0 ;
int begZin = 0;
int end1 = 0;
int klMax1 = 0;

import controlP5.*;
ControlP5 controlP5;



String textValue = "";
Textfield myTextfield;

void setup() {
  size(400,400);
  frameRate(25);
  controlP5 = new ControlP5(this);

  myTextfield = controlP5.addTextfield("texting",20,20,200,20);
  myTextfield.setFocus(true);



}

void draw() {
  background(0);


}





public void texting(String theText) {
  // receiving text from controller texting
  println(theText);
  //  String str = theText
  char[] theText2 = theText.toCharArray();
  println (theText2);

  for (int i=0; i < theText2.length; i++) {
    if (theText2[i] == ' '){ 
	// spcAr[x] = i; 
	spc.add(new Integer(i));



    }


  }


  //println(theText2.length);
 // println("array waarde 1 = "+spc.get(0));
  //println("array waarde 2 = "+spc.get(1));

/*if (spc.get(0) == " "){
  int klMax1 = (Integer)spc.get(0)-1;
}
else {
  int klMax1 = theText2.length;
}
*/
  int klMax1 = (Integer)spc.get(0)-1;
 

  for (int kl1=0; kl1 < (klMax1)+1; kl1++) {
   // println(kl1);
    if (theText2[kl1] == 'a'){ 
	WDstA1.add('a'); 
	WDnrA1.add(kl1);	
	int  WDnr1 = i;
    }
    else if(theText2[kl1] == 'o'){ 
	WDstA1.add('o'); 
	WDnrA1.add(kl1);	
	int WDnr1 = i;
    }
    else if(theText2[kl1] == 'e'){ 
	WDstA1.add('e');
	WDnrA1.add(kl1); 	
	int WDnr1 = i;
    }
    else if(theText2[kl1] == 'u'){ 
	WDstA1.add('u');
	WDnrA1.add(kl1); 	
	int WDnr1 = i;
    }
    else if(theText2[kl1] == 'i'){ 
	WDstA1.add('i');
	WDnrA1.add(kl1); 	
	int WDnr1 = i;
    }
    else {
	int WDnr1 = i;
    }
	



  }
  println ("woord klinker(s) = "+ WDstA1);
  println ("Positie klinker = "+  WDnrA1 );

  if ( (Integer)WDnrA1.get(0) == 0 ) {
    int begZin = 1;
    println("begin zin is wel een klinker");
  }
  //else if ( WDnrA1.get(0) == 0
  //////////////////////////////////////////////////////// bekijkt hoeveelheid klinkers en weergeeft ze  van achteren naar voren
  for (int x=0; x < (Integer)WDstA1.size(); x++){
    println(WDnrA1.get(x) + "array nummer " +WDstA1.get(x) + " letter " + " potitie " +x);
   
    if ((WDnrA1.contains(x) == true) && (WDnrA1.contains(x+1) == true) && (WDnrA1.contains(x+2) == true)
	   && (WDnrA1.contains(x+3) == true) && (WDnrA1.contains(x+4) == true)){
	println ("vijf klinkers naast elkaar = "+ WDstA1.get(x)+ WDstA1.get(x+1) + WDstA1.get(x+2) + WDstA1.get(x+3) + 
		    WDstA1.get(x+4));
	// x = (Integer)WDstA1.size();
		
    }
     else if ((WDnrA1.contains(x) == true) && (WDnrA1.contains(x+1) == true) && (WDnrA1.contains(x+2) == true)  
	   && (WDnrA1.contains(x = +3) == true)){
	println ("vier klinkers naast elkaar = "+ WDstA1.get(x)+ WDstA1.get(x+1) + WDstA1.get(x+2) + WDstA1.get(x+3));
	 //x = (Integer)WDstA1.size();
	     
	  
	   }
    else if ((WDnrA1.contains(x) == true) && (WDnrA1.contains(x+1) == true) && (WDnrA1.contains(x+2) == true)){
	println ("drie klinkers naast elkaar = "+ WDstA1.get(x)+ WDstA1.get(x+1) + WDstA1.get(x+2));
	// x = (Integer)WDstA1.size();
	 
		  
    }
    else if ((WDnrA1.contains(x) == true) && (WDnrA1.contains(x+1) == true)){
	println ("twee klinkers naast elkaar = "+ WDstA1.get(x)+ WDstA1.get(x+1));
	// x = (Integer)WDstA1.size();
	    
		  
    }
    else if ((WDnrA1.contains(x) == true)){
	println ("een klinkers naast elkaar = "+ WDstA1.get(x));
	// x = (Integer)WDstA1.size();
	
	 //x = (Integer)WDstA1.size();
    }
 
  }
   WDstA1.clear();
  WDnrA1.clear();
    }

  

 

At first i want to search the word for the first vowel and possibly second and third or fourth. After i know where the vowels are i want to add consonants possibly before and possibly after the vowel. I allready have a database with the cutted parts of the words. This is like the basic idea of what i want to do. It should be simple. I want to do everything into arrays, because i understand that, but i only get the space checking between the words in an arraylist. and when i want to ask the first value of the arraylist, i get two different values.

It is in Programs because the first messages were about program design, not syntax.

Some remarks on your previous message: the int  WDnr1 = i; lines in tests are useless because the variable you create there disappears after the closing brace.
Create int WDnr1 = -1; before the test and assign like: WDr1 = i; (well, if you plan to use it).
And, well, I have not understood if you have a question there.

On the last message, the [I@45fe3b is the result we get when we do a System.out.println(spcAr) when spcAr is an array of int. But Processing is supposed to print that OK. Ah, no, it works only with println(), not with print().

No need to fully quote the message which is just above your...

I don't understand the purpose of your tests WDnrA1.contains(x+3) (you can skip the == true too).
Apparently, you put chars in the array list, and you go through the whole length of this list to
Wait, no, that's WDstA1, the similar names are confusing.

But your algorithm seems complex. Can't you use String.indexOf function instead? Or do I miss something (haven't tried hard to understand it).