Java:如何检查对角线连接二维阵列中的四连胜

问题描述

我有一个连接四板"，这是一个6 * 7 2D字符数组，其中填充有X或O两个空格.当垂直，水平或垂直连续四个X或四个O时，满足获胜条件.对角地.我设法成功检查了垂直和水平方向的获胜条件，其中通过以下方法返回了获胜的字符:

I have a Connect Four "board" which is a 6*7 2D char array populated with either spaces, X or O. The win condition is met when there are either four Xs or four Os in a row vertically, horizontally or diagonally. I've managed to get the win conditions checked successfully for vertical and horizontal, where the winning char is returned by some methods as below:

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;
}

我只是停留在如何检查对角线胜利的问题上，而不会抛出ArrayIndexOutOfBoundsException.我知道我需要遍历2D数组两次，一次是对前对角线，一次是对后对角线 4平方或更多，如下图所示:

I'm just stuck on how to check for diagonal wins, without throwing an ArrayIndexOutOfBoundsException. I know I need to iterate through the 2D array twice, once for forward diagonals and once for backward diagonals that are 4 squares long or more, like in the below diagram:

对角线要检查的图

基本上，我该如何填写此方法以返回获胜的字符?

Basically, how would I fill in this method to return a winning char?

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

    // some iteration here

    return dWinner;
}

任何帮助将不胜感激！

推荐答案

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

使用代码:

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
}

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