Question

我是android开发的初学者，我正在尝试通过讲师指导的课程来制作井字游戏。

我已经为两个玩家添加了两个获胜条件，但是我想为没有获胜者或平局添加条件。

我该怎么做？

这是我的代码：

public class MainActivity extends AppCompatActivity {

    // 0 : yellow, 1 : red, empty : 2

    int[] gameState = {2,2,2,2,2,2,2,2,2,};

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

    int activePlayer = 0;

    boolean gameActive = true;

    public void dropIn(View view) {
        ImageView counter = (ImageView) view;

        int tappedCounter = Integer.parseInt(counter.getTag().toString());

        if (gameState[tappedCounter] == 2 && gameActive) {

            gameState[tappedCounter] = activePlayer;

            counter.setTranslationY(-1500);
            if (activePlayer == 0) {

                counter.setImageResource(R.drawable.yellow);
                activePlayer = 1;

            } else {

                counter.setImageResource(R.drawable.red);
                activePlayer = 0;
            }
            counter.animate().translationYBy(1500).setDuration(400);
            for (int[] winningPosition : winningPositions) {
                if (gameState[winningPosition[0]] == gameState[winningPosition[1]] && gameState[winningPosition[1]] == gameState[winningPosition[2]] && gameState[winningPosition[0]] != 2) {

                    String winner = "";

                    gameActive = false;
                    if (activePlayer == 1) {
                        winner = "Yellow";
                    } else {
                        winner = "Red";
                    }
                    //Toast.makeText(this, winner + " have won!", Toast.LENGTH_SHORT).show();

                    Button playAgainButton = (Button) findViewById(R.id.playAgainButton);
                    TextView winnerTextView = (TextView) findViewById(R.id.winnerTextView);

                    winnerTextView.setText(winner + " has won!");

                    playAgainButton.setVisibility(View.VISIBLE);
                    winnerTextView.setVisibility(View.VISIBLE);


                }
            }
        }
    }

    public void playAgain(View view){
        Button playAgainButton = (Button) findViewById(R.id.playAgainButton);
        TextView winnerTextView = (TextView) findViewById(R.id.winnerTextView);
        playAgainButton.setVisibility(View.INVISIBLE);
        winnerTextView.setVisibility(View.INVISIBLE);

        GridLayout gridLayout = findViewById(R.id.gridLayout);
        for(int i=0; i < gridLayout.getChildCount(); i++){
            ImageView counter = (ImageView) gridLayout.getChildAt(i);
            counter.setImageDrawable(null);
        }
        for(int i = 0; i < gameState.length; i++){
            gameState[i] = 2;
        }

        activePlayer = 0;

        gameActive = true;
    }

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
    }
}

Answer 1

这非常简单-您必须知道何时获取每个字段（有人在上面加上“ X”或“ O”）因此，当每个领域都被占用并且没有获得任何获胜位置时-这就是平局。

从程序的角度来看：

1）当玩家1或玩家2进行移动时-计算剩下的可用字段

2）当自由字段== 0时，调用draw（）函数或类似的东西

Answer 2

没有条件，没有赢家，因为转牌由你决定，规则相同。

您可以使用Minimax algorithm作弊来预测玩家的下一轮。如果他们相信算法，那么只有玩家才能赢。

对于Case Player和计算机，Minimax算法也是一个不错的选择。您无法击败计算机。

此外，alpha beta algorithm得到了改善。

您可以通过Java Here来引用tictactoe窗口应用程序

井字游戏没有赢家

2 个答案: