Question

我正在创建一个由JButtons组成的二维单元格数组（如果有意义的话）来创建一个类似于井字游戏的游戏。 Example of what im trying to accomplish 忘了添加，如果一个小区被包围，则赢得游戏

我目前在编辑for循环时失去了考虑从当前单元格中分配一个或零，并将其分配给局部变量。

当我说topNum = board [row-1] [col]; ，我将如何检索分配给该单元格的一个或零。（我使用setText属性将数字分配给单元格。）

感谢您的帮助

public boolean isWinner() {
    for (int row = 0; row < this.sizeRow; row++) {
        for (int col = 0; col < this.sizeCol; col++) {
            Cell num = board[row][col];

            // not a winner if this cell is null
            if (num != null) {

                // Look at above cell
                Integer topNum = -1;
                if (row > 0) {
                    topNum = board[row-1][col];
                    if (topNum == null) break;
                }                 

                // Look at right cell
                Integer rightNum = -1;
                if (col == this.sizeCol-1) {
                    rightNum = board[row][col+];
                    if (rightNum == null) break;
                }

                // Look at left cell
                Integer leftNum = -1;
                if (col == this.sizeCol-1) {
                    rightNum = board[row][col-];
                    if (rightNum == null) break;
                }

                // Look at bottom cell
                Integer bottomNum = -1;
                if (row > 0) {
                    topNum = board[row+1][col];
                    if (topNum == null) break;
                }  


                // do similar for left cell and bottom cell

                // Check that topNum is not our current cell
                // then check that topNum is the same number as right left and bottom OR those cells are -1, which
                // means that side was on an edge.
                if (!num.equals(topNum)) {
                    if ((topNum.equals(rightNum) || rightNum.equals(-1))
                      && (topNum.equals(leftNum) || leftNum.equals(-1))
                      && (topNum.equals(bottomNum) || bottomNum.equals(-1)) {
                        return true;
                    }
                }
            }
        }
    }
    return false;
}

Answer 1

从简单的事情开始，确定一个单元格当前是否被玩家占用

protected boolean isOffBoard(int x, int y) {
    return x < 0 || y < 0 || y >= board.length || x >= board[y].length;
}

protected boolean isOccupied(int x, int y) {
    if (isOffBoard(x, y)) {
        return true;
    }

    return board[y][x] != null;
}

protected boolean isOccupiedByPlayer(int player, int x, int y) {
    if (isOffBoard(x, y)) {
        return true;
    }
    if (!isOccupied(x, y)) {
        return false;
    }

    return board[y][x] == player;
}

这三种方法做得很简单

确定电池是否不在电路板上
确定单元格是否被占用
确定单元格是否被特定玩家占用

这很方便，因为现在我们可以简单地对特定点进行循环检查，并确定玩家是否被包围，这样的事情......

protected boolean isSurroundedBy(int player, int x, int y) {
    return isOccupiedByPlayer(player, x - 1, y - 1)
            && isOccupiedByPlayer(player, x, y - 1)
            && isOccupiedByPlayer(player, x + 1, y - 1)
            && isOccupiedByPlayer(player, x - 1, y)
            && isOccupiedByPlayer(player, x + 1, y)
            && isOccupiedByPlayer(player, x - 1, y + 1)
            && isOccupiedByPlayer(player, x, y + 1)
            && isOccupiedByPlayer(player, x + 1, y + 1);
}

这只是检查给定点周围的单元格是否被特定玩家占用

我使用以下方法测试它......

board[1][1] = 0;
System.out.println("");
System.out.println("1x1 isOccupiedByPlayer 0 = " + isOccupiedByPlayer(0, 1, 1));
System.out.println("isSurrounded = " + isSurroundedBy(1, 1, 1));
board = new Integer[3][3];
board[0][0] = 0;
System.out.println("");
System.out.println("0x0 isOccupiedByPlayer 0 = " + isOccupiedByPlayer(0, 0, 0));
System.out.println("isSurrounded = " + isSurroundedBy(1, 0, 0));
board[1][0] = 1;
board[1][1] = 1;
board[0][1] = 1;
System.out.println("");
System.out.println("0x0 isOccupiedByPlayer 0 = " + isOccupiedByPlayer(0, 0, 0));
System.out.println("1x1 isOccupiedByPlayer 0 = " + isOccupiedByPlayer(0, 1, 1));
System.out.println("isSurrounded = " + isSurroundedBy(1, 0, 0));

board = new Integer[3][3];
board[0][0] = 0;
board[1][0] = 1;
board[1][1] = 0;
board[0][1] = 1;
System.out.println("");
System.out.println("0x0 isOccupiedByPlayer 0 = " + isOccupiedByPlayer(0, 0, 0));
System.out.println("1x1 isOccupiedByPlayer 0 = " + isOccupiedByPlayer(0, 1, 1));
System.out.println("isSurrounded = " + isSurroundedBy(1, 0, 0));

哪些输出......

1x1 isOccupiedByPlayer 0 = true
isSurrounded = false

0x0 isOccupiedByPlayer 0 = true
isSurrounded = false

0x0 isOccupiedByPlayer 0 = true
1x1 isOccupiedByPlayer 0 = false
isSurrounded = true

0x0 isOccupiedByPlayer 0 = true
1x1 isOccupiedByPlayer 0 = true
isSurrounded = false

现在，这可能无法满足您的绝对需求，但应该给您一个跳跃点

Answer 2

如果您使用setText（）将一个或零分配给单元格。我会使用getText（）来检索指定的值。

使用二维数组创建游戏

2 个答案: