Question

我在Java中创建了一个小TicTacToe游戏，我想编写更有效的代码，我会创建一个for循环来创建9个按钮。

我遇到的问题是弄清楚如何测试现在按哪个按钮以确定胜利者。我已经注释掉了我的旧测试代码，因为它不再有效。

import java.awt.*;
import java.awt.event.*;

import javax.swing.*;

public class TicTacToe extends JFrame implements ActionListener, WindowListener{

public JFrame window = new JFrame("window");



public String letter = "";
public String lastLetter = "";
public int count = 0;
public boolean win;
public boolean isTieGame;
public static int x;


public TicTacToe()
{

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

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


    window.setVisible(true);
    }

public static void main(String[] args) {
    new TicTacToe();
    }

public void counter()
    {
        if(count == 1 || count == 3 || count == 5 || count == 7 || count == 9)
        {
            letter = "X";
        }
        else
        {
            letter = "O";
        }
    }
/*
public boolean isGameOver()
{

    if(b1.getText().equals(b2.getText()) && b2.getText().equals(b3.getText()) && b1.getText() != "")
    {
        win = true;
    }
    else if(b4.getText().equals(b5.getText()) && b5.getText().equals(b6.getText()) && b4.getText() != "")
    {
        win = true;
    }
    else if(b7.getText().equals(b8.getText()) && b8.getText().equals(b9.getText()) && b7.getText() != "")
    {
        win = true;
    }
    else if(b1.getText().equals(b5.getText()) && b5.getText().equals(b9.getText()) && b1.getText() != "")
    {
        win = true;
    }
    else if(b3.getText().equals(b5.getText()) && b5.getText().equals(b7.getText()) && b3.getText() != "")
    {
        win = true;
    }
    else if(b1.getText().equals(b4.getText()) && b4.getText().equals(b7.getText()) && b1.getText() != "")
    {
        win = true;
    }
    else if(b2.getText().equals(b5.getText()) && b5.getText().equals(b8.getText()) && b2.getText() != "")
    {
        win = true;
    }
    else if(b3.getText().equals(b6.getText()) && b6.getText().equals(b9.getText()) && b3.getText() != "" )
    {
        win = true;
    }
    else
    {
        win = false;
    }
    return win;
}
*/
public void endOrReset()
    {
        if(win)
        {
            JOptionPane.showMessageDialog(null, lastLetter() + " WINS!");
            int playAgain = JOptionPane.showConfirmDialog(null, "Would you like to play again?", "Play Again.", JOptionPane.YES_NO_OPTION);
            if(playAgain == (JOptionPane.YES_OPTION))
            {
            win = false;
            }
            else if(playAgain == JOptionPane.NO_OPTION)
            {
                JOptionPane.showMessageDialog(null, "Goodbye");
                System.exit(0);
            }
        }
        else if(isTieGame)
            {
            JOptionPane.showMessageDialog(null, "Tie Game!");
            int playAgain = JOptionPane.showConfirmDialog(null, "Would you like to play again?", "Play Again.", JOptionPane.YES_NO_OPTION);
            if(playAgain == JOptionPane.YES_OPTION)
            {


            }
            else if(playAgain == JOptionPane.NO_OPTION)
            {
                JOptionPane.showMessageDialog(null, "Goodbye");
                System.exit(0);
            }
            }

    }
public void actionPerformed(ActionEvent a) {

    count++;
    counter();
    //isGameOver();
    endOrReset();



    ((JButton)a.getSource()).setText(letter);

   }

@Override
public void windowActivated(WindowEvent e) {
    // TODO Auto-generated method stub

}

@Override
public void windowClosed(WindowEvent e) {
    // TODO Auto-generated method stub

}

@Override
public void windowClosing(WindowEvent e) {
    // TODO Auto-generated method stub

}

@Override
public void windowDeactivated(WindowEvent e) {
    // TODO Auto-generated method stub

}

@Override
public void windowDeiconified(WindowEvent e) {
    // TODO Auto-generated method stub

}

@Override
public void windowIconified(WindowEvent e) {
    // TODO Auto-generated method stub

}

@Override
public void windowOpened(WindowEvent e) {
    // TODO Auto-generated method stub

}   
public String lastLetter()
{
    String lastLetter;
    if(letter == "O")
    {
        lastLetter = "X";
    }
    else
    {
        lastLetter = "O";
    }
    return lastLetter;
}
public boolean isTieGame()
{
    if(count >= 9 && win == false)
    {
        isTieGame = true;
    }
    return isTieGame();
}
}

Answer 1

我认为最简单的解决方案是将JButton[] b设为字段，因此将其声明在构造函数上方。

在这种情况下，您还应该使用更具描述性的名称。

或者你可以拿着一个char数组，其中一个按钮记录当前的玩家符号（X或O），这就是我前一段时间解决这个问题的方法

Answer 2

我喜欢做的是拥有一个独立于GUI的9长度数组来保持游戏状态，如下所示：

0 1 2
3 4 5
6 7 8

如果您考虑指数，

3*(index/3) + index % 3

这通常是一个枚举数组，但你也可以使用String数组，

public enum State
{
    EMPTY,
    X,
    O;
}

然后

State[] gridState = new State[9];
for(int i = 0; i < gridState.length; i++) {
    gridState = State.EMPTY;
}

如果您根据指标根据GUI初始化按钮，

public class MyButton extends JButton {
    private int index;

    public MyButton(int index) {
        super();
        this.index = index;
    }

    public int getIndex() {
        return index;
    }
}

然后当有人点击你的按钮时，你可以说（（myButton的）a.getSource（））。getIndex（）

您将获得这个按钮的索引。

如果tic tac toe被定义为3x3和2D，我喜欢根据索引来检查胜利，我将存储一组数组，这些数组包含胜利的水平，垂直和对角线索引。

int[][] winIndices = {{0,1,2}, {3,4,5}, {6,7,8}, {0,3,6}, {1,4,7}, {2,5,8}, {0,4,8}, {2,4,6}};

然后你可以根据

检查胜利

public boolean checkWin() {
    for(int[] indexArray : winIndices) {
        boolean same = (gridState[indexArray[0]] == gridState[indexArray[1]]) && 
                       (gridState[indexArray[1]] == gridState[indexArray[2]])
        if(same == true && gridState[indexArray[0]] != State.EMPTY) {
             return true;
        }
    }
    return false;
}

或类似的东西。我曾经把它写得更好了，但这不是在这台电脑上，但它是类似的东西：）

Tic-Tac-Toe Java：按钮问题

2 个答案: