10 minute Tic Tac Toe

Question

After seeing Snake coded in 10 minutes, I decided to try a similar challenge, albeit easier. The following is Tic Tac Toe that I coded in 10 minutes. Given the short time frame, I'm sure there are style and convention errors abound, and would appreciate any suggestions :)

I will also specifically ask about one thing - checking if there is a winner(done in the only ActionPerformed method). It feels long and convoluted, but it was the best I could think of at the time. Looking back, I still can't really think of any ways to write it better, actually...

import javax.swing.*;
import java.awt.event.*;
import java.awt.*;
public class TicTacToe {
  private JFrame main;
  private TicButton[][] buttons;
  private JPanel[] rows;
  private boolean p1turn = true;
  private final Color bg1 = new Color(255,0,0);
  private final Color bg2 = new Color(0,255,0);
  public static void main(String[] args) {
    new TicTacToe().go();
  }
  public void go() {
    main = new JFrame("Tic Tac Toe!");
    buttons = new TicButton[3][3];
    rows = new JPanel[3];
    int r = 0;
    main.getContentPane().setLayout(new BoxLayout(main.getContentPane(),BoxLayout.Y_AXIS));
    for (int i = 0; i < 3; i++) {
      rows[r] = new JPanel();
      for (int j = 0; j < 3; j++) {
        buttons[i][j] = new TicButton();
        rows[r].add(buttons[i][j]);
      }
      main.getContentPane().add(rows[r]);
      r++;
    }
    main.pack();
    main.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    main.setVisible(true);
  }
  private class TicButton extends JButton{
    int value;
    private boolean clicked;
    public TicButton() {
      super("-");
      this.value = -1;
      clicked = false;
      setFocusable(false);
      this.addActionListener(new ActionListener(){  
        public void actionPerformed(ActionEvent e){  
            if (!clicked) {
              clicked = true;
              setText(p1turn ? "X" : "O");
              value = p1turn ? 1 : 2;
              setBackground(p1turn ? bg1 : bg2);
              p1turn = !p1turn;
              setEnabled(false);
            }

          boolean winner = false;
          String win = "";
          for (TicButton[] tt : buttons) {
            if (tt[0].value == tt[1].value && tt[0].value == tt[2].value && tt[0].value != -1) {
              winner = true;
              win = tt[0].value == 1 ? "X" : "O";
              break;
            }
          }
          for (int j = 0; j < 3; j++) {
            if (buttons[0][j].value == buttons[1][j].value && 
                buttons[0][j].value == buttons[2][j].value &&
                buttons[0][j].value != -1) {
              winner = true;
              win = buttons[0][j].value == 1 ? "X" : "O";
              break;
            }
          }
          if (buttons[0][0].value == buttons[1][1].value && buttons[0][0].value == buttons[2][2].value &&
              buttons[0][0].value != -1) {
            winner = true;
            win = buttons[0][0].value == 1 ? "X" : "O";
          }
          if (buttons[0][2].value == buttons[1][1].value && buttons[0][2].value == buttons[2][0].value &&
              buttons[0][2].value != -1) {
            winner = true;
            win = buttons[0][2].value == 1 ? "X" : "O";
          }
          if (winner) {
            JOptionPane.showMessageDialog(main, win + " wins!");
            main.setVisible(false);
            System.exit(0);
          }
        }  
      }); 
    }
  }

}

Answer 1

When it is X’s turn, and they make a move, only player X can win. Similarly, only Y can win on Y’s turn. So why all these tests?

win = tt[0].value == 1 ? "X" : "O";

If there is a winner, just test p1turn in the if (winner) { code!

String win = p1turn ? "X" : "O";

Okay, you may want to move p1turn = !p1turn; to the end of the function, first, to avoid X/O confusion.

---

This code is checking if the row has all the same symbols, and the symbol is not blank:

if (tt[0].value == tt[1].value && tt[0].value == tt[2].value && tt[0].value != -1) {

When the player moves, this.value is set to their code. And only they can win on this move. So we could instead write:

if (tt[0].value == value && tt[1].value == value && tt[2].value == value) {

Slightly shorter. But we can do better.

Change X’s & O’s values to +1 and -1. Use 0 for blank. Then, if a row sums to +3 or -3, you have a winner!

if (tt[0].value + tt[1].value + tt[2].value == 3*value) {

Similarly for other winning directions.