Java:如何检查二维数组中的对角线连接四

Question

我有一个 Connect Four “board”，它是一个 6*7 二维字符数组，其中填充了空格、X 或 O。当垂直、水平或连续的一行中有四个 X 或四个 O 时，满足获胜条件对角地。我已经成功地检查了垂直和水平的获胜条件，其中获胜字符通过以下一些方法返回:

private char CheckVerticalWinner(char[][] currentBoard) {
    // check vertical (move down one row, same column)
    char vWinner = ' ';
    for (int col=0; col<7; col++) {
        int vCount = 0;
        for (int row=0; row<5; row++) {
            if (currentBoard[row][col]!=' ' &&
                currentBoard[row][col] == currentBoard[row+1][col]) {
                vCount++;
                System.out.println("VERT "+vCount); //test
            } else {
                vCount = 1;
            }

            if (vCount>=4) {
                vWinner = currentBoard[row][col];
            }
        }
    }
    return vWinner;
}


private char CheckHorizontalWinner(char[][] currentBoard) {
    // check horizontal (move across one column, same row)
    char hWinner = ' ';
    for (int row=0; row<6; row++) {
        int hCount = 0;
        for (int col=0; col<6; col++) {
            if (currentBoard[row][col]!=' ' &&
                currentBoard[row][col] == currentBoard[row][col+1]) {
                hCount++;
                System.out.println("HORIZ "+hCount); //test
            } else {
                hCount = 1;
            }

            if (hCount>= 4) {
                hWinner = currentBoard[row][col];
            }
        }
    }
    return hWinner;
}

我只是想知道如何在不抛出 ArrayIndexOutOfBoundsException 的情况下检查对角线胜利。我知道我需要遍历二维数组两次，一次用于前向对角线，一次用于长度为 4 个正方形或更多的后向对角线，如下图所示:

Diagonals to be checked diagram

基本上，我将如何填写此方法以返回获胜字符？

private char CheckDiagonalWinner(char[][] currentBoard) {

    // some iteration here

    return dWinner;
}

如有任何帮助，我们将不胜感激!

Answer 1

最简单的算法可能是:

for every direction
    for every coordinate
        check whether the next 3 elements in this direction exist and are the same

在代码中:

final int maxx = 7;
final int maxy = 6;

char winner(char[][] board) {
    int[][] directions = {{1,0}, {1,-1}, {1,1}, {0,1}};
    for (int[] d : directions) {
        int dx = d[0];
        int dy = d[1];
        for (int x = 0; x < maxx; x++) {
            for (int y = 0; y < maxy; y++) {
                int lastx = x + 3*dx;
                int lasty = y + 3*dy;
                if (0 <= lastx && lastx < maxx && 0 <= lasty && lasty < maxy) {
                    char w = board[x][y];
                    if (w != ' ' && w == board[x+dx][y+dy] 
                                 && w == board[x+2*dx][y+2*dy] 
                                 && w == board[lastx][lasty]) {
                        return w;
                    }
                }
            }
        }
    }
    return ' '; // no winner
}