TicTacToe game needs optimizing

Question

I'm new to programming and want to check if I'm using best coding practices in my program.

package scott.TicTacToe.main;

import java.awt.GridLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;

import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JOptionPane;

public class TicTacToe implements ActionListener{
    private JFrame window = new JFrame("Tic-Tac-Toe");
    private JButton[] buttons = new JButton[9];
    private String letter = "";
    private int count = 0;
    private boolean win = false;

    public TicTacToe(){
        window.setSize(300, 300);
        window.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        window.setLayout(new GridLayout(3, 3));

        for(int i = 0; i < 9; i++){
            buttons[i] = new JButton();
            window.add(buttons[i]);
            buttons[i].addActionListener(this);
        }

        window.setVisible(true);

    }

    public void actionPerformed(ActionEvent event) {

        count++;
        if(count % 2 != 0){
            letter = "X";
        } else {
            letter = "O";
        }

        for(int i = 0; i < 9; i++){
            if (event.getSource() == buttons[i]){
                buttons[i].setText(letter);
                buttons[i].setEnabled(false);
            }

            //Horizontal Wins
            if(buttons[0].getText().equals(buttons[1].getText()) && buttons[1].getText().equals(buttons[2].getText()) && !buttons[0].getText().equals("")){
                win = true;
            } else if (buttons[3].getText().equals(buttons[4].getText()) && buttons[4].getText().equals(buttons[5].getText()) && !buttons[3].getText().equals("")){
                win = true;
            } else if (buttons[6].getText().equals(buttons[7].getText()) && buttons[7].getText().equals(buttons[8].getText()) && !buttons[6].getText().equals("")){
                win = true;
            }

            //Vertical Wins
            else if(buttons[0].getText().equals(buttons[3].getText()) && buttons[3].getText().equals(buttons[6].getText()) && !buttons[0].getText().equals("")){
                win = true;
            } else if(buttons[1].getText().equals(buttons[4].getText()) && buttons[4].getText().equals(buttons[7].getText()) && !buttons[1].getText().equals("")){
                win = true;
            } else if(buttons[2].getText().equals(buttons[5].getText()) && buttons[5].getText().equals(buttons[8].getText()) && !buttons[2].getText().equals("")){
                win = true;
            }

            //Diagonal Wins
            else if(buttons[0].getText().equals(buttons[4].getText()) && buttons[4].getText().equals(buttons[8].getText()) && !buttons[0].getText().equals("")){
                win = true;
            } else if(buttons[6].getText().equals(buttons[4].getText()) && buttons[4].getText().equals(buttons[2].getText()) && !buttons[6].getText().equals("")){
                win = true;
            } else {
                win = false;
            }

            if(win == true){
                JOptionPane.showMessageDialog(null, letter + " wins!");
            } else if(count == 9){
                JOptionPane.showMessageDialog(null, "It's a draw!");
                count = 0;
            }
        }

    }

    public static void main(String[] args) {

        new TicTacToe();

    }

}

Answer 1

You could perhaps look at separating the win condition by sticking it in another method then call it in ActionPerformed.

Outside of that, you could replace:

if(count % 2 != 0){
        letter = "X";
    } else {
        letter = "O";
    }

with a ternary operator such as:

letter =(count%2!=0)?"X":"O";

That's all I can think off the top of my head.