how to do a buble sort animation using java?

That is definitely not the loop for bubble-sort, and I still don't understand your question properly. Do you want to first display the unsorted array, then sort it, and then display the sorted array again?

Mayank.

Hi mayankeagle,

My code is correct but not efficient like yrs, given the time complexity of bubble sort is O(n^2). So logically I thought it would go something like that.

You're right.

Anyway, I just wrote a little code for the animation here using your version. How creative are you???

public class BubbleSortAnimation extends Frame{
  int[] data;
  Button[] b_data;
  final int MAX =10;
  Panel p;
  Label l;

  public BubbleSortAnimation() {
    data = new int[MAX];
    b_data = new Button[MAX];
    p = new Panel(new FlowLayout());
    Random r = new Random();
    int temp;
    for(int i=0;i<MAX;i++){
      temp = r.nextInt(100);
      data[i]=temp;
      b_data[i] = new Button(Integer.toString(temp));
      b_data[i].setEnabled(false);
      b_data[i].setBackground(Color.white);
      p.add(b_data[i]);
    }
    l = new Label();
    this.add("Center",p);
    this.add("South",l);
    this.setSize(600,100);
    this.setVisible(true);
  }

  public void bubbleSort() throws Exception{
    l.setText("Sorting....");
    repaint();
    for(int i=1;i<MAX;i++)
      for(int j=0;j<MAX-i;j++){
         b_data[j].setBackground(Color.green);
         b_data[j+1].setBackground(Color.yellow);
         repaint();
         Thread.sleep(500);
         if(data[j]>data[j+1]){
           int temp = data[j];
           data[j] = data[j+1];
           data[j+1] = temp;
           b_data[j].setBackground(Color.yellow);
           b_data[j].setLabel(Integer.toString(data[j]));
           b_data[j+1].setBackground(Color.green);
           b_data[j+1].setLabel(Integer.toString(data[j+1]));
           repaint();
           Thread.sleep(500);
         }
         b_data[j].setBackground(Color.white);
         b_data[j+1].setBackground(Color.white);
         Thread.sleep(500);
      }
    l.setText("Sorted!!!");
    repaint();
  }

  public static void main(String[] args) {
    BubbleSortAnimation b = new BubbleSortAnimation();
    try{
      b.bubbleSort();
    }
    catch(Exception e){
      System.out.println(e);
    }
  }
}

Bye.

His code is correct.  Remember C arrays index from 0 to Length-1  So Arr[6] has indexes 0, 1, 2 , 3 , 4 , 5

The code provided by funnyveryfunny and mayankeagle will step outside the array boundaries when y = length-1

mlmcc

To mlmcc,

>>>"The code provided by funnyveryfunny and mayankeagle will step outside the array boundaries when y = length-1"

First of all your statement is correct in its respect but I think you're confused here, y only reaches y = length-2 so it'll never step outside the upper-boundary. To support yr statement the code will now have to be like this:

for(int x = 0; x <= array.length - 1; x++){
   for(int y = 0; y <= array.length - 2; y ++){
        if(array[y] > array[y+1]){
           swap(array, y, y+1);
        }
    }
}

Otherwise the last element will never be sorted unless yr assuming item[y]<=item[length-1] for all y=0..length-2.

fvf.

mlmcc ,

>> The code provided by funnyveryfunny and mayankeagle will step outside the array boundaries when y = length-1

Oh ya??

WELL, 'y' WILL NEVER BE 'LENGTH - 1' IF YOU FOLLOW MY CODE,

because 'x' starts from 1 in my loop and goes till n - 1 (n-2 iterations for the outer loop - standard bubble-sort), and so, the maximum value of 'y' ( y < array.length - x ) will be when 'x' is 1, i.e., 'array.length - 2', meaning that (y + 1) in the 'if' statement will be array.length - 1, and so, both, 'y' and 'y + 1' will be within the array's bounds!

Please verify your own comments before posting them and contradicting somebody else.

Mayank.

tttntt:
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Split points between mayankeagle and funnyveryfunny.

Ok with me.

Me too :-)