Question

我试图创建Tic Tac Toe，我能够填满我的董事会，我能够检查谁获胜的行和列。但是，我需要一些帮助来对角检查，看谁赢了。这就是我到目前为止所拥有的。我是初学者，所以请不要过于努力。

检查方法：

public boolean check(String[] [] board)
{
    int j = 0;
    for(int i = 0; i < board.length;i++) //row
    {
        if(board[i][j] == "X" || board[i][j+1] == "X" || board[i][j+2] == "X")
        {
          if(board[i][j] == board[i][j+1] & board[i][j] == board[i][j+2])
          {
              System.out.println("X wins!");
                return true;
          }
        }
    }
    int e = 0;
    for(j = 0; j < board.length;j++) //col
    {
        if(board[e][j] == "X" | board[e + 1][j] == "X"  | board[e + 2][j] == "X" )
        {
          if(board[e][j] == board[e + 1][j] & board[e][j] == board[e + 2][j])
          {
              System.out.println("X, wins");
                return true;
          }
        }
     }
    int d = 0;
    for(int ii = 0; ii < board.length; ii++) //diag
    {
        if(board[d][ii] != null || board[d + 1][ii] != null | board[d + 2][ii] != null)
        {
            if(board[d][ii] == board[d + 1][ii+1] & board[d][ii] == board[d + 2][ii])
            {
                System.out.println("X, wins dig");
                return true;
            }
        }
    }
    int k = 0;
    for(int i = 0; i < board.length;i++) //row
    {
        if(board[i][k] == "O" || board[i][k+1] == "O" || board[i][k+2] == "O")
        {
          if(board[i][k] == board[i][k+1] & board[i][k] == board[i][k+2])
          {
              System.out.println("O wins!");
                return true;
          }
        }
    }
    int z = 0;
    for(k = 0; k < board.length; k++) //col
    {
        if(board[z][k] == "O" | board[z + 1][k] == "O"  | board[z + 2][k] == "O" )
        {
          if(board[z][k] == board[z + 1][k] & board[z][k] == board[z + 2][k])
          {
              System.out.println("O, wins");
                return true;
          }
        }
     }

    return false;
}

Answer 1

这个问题有一些你想要做的事情：

首先确保价值已经不存在:(评论中的建议考虑在内）

public void fillX(String[][] board) {
    attemptToFillValue(board, "x");
}

public void fillO(String[][] board) {
    attemptToFillValue(board, "o");
}

private void attemptToFillValue(String[][]board,  String value) {
    do {
        int r = Integer.parseInt(JOptionPane.showInputDialog("What row from 1-3?"));
        int c = Integer.parseInt(JOptionPane.showInputDialog("What column from 1-3?"));
        if (board[r - 1][c - 1] != null) {
            board[r - 1][c - 1] = value;
            return;
        } else {
            //Warn then that value already exists in some way
        }
    } while (true);
}

关于验证董事会是否有赢家，我建议使用for循环并进行3次单独检查。水平，垂直和对角线。

我不想为你做整个任务，所以留给你

Tic Tac Toe - 2D阵列

1 个答案: