为我的Java方格棋盘游戏优化我的“捕获”算法

Question

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• ex)白条在板子的左边缘，然后黑条直接移动到白条的右边，这将是一个捕获

到目前为止我有一个

• 11 x 11矩阵
• （int）0=空，1=白，2=黑，3=白

但是我不能这么简单地做，因为如果这个片段在两个外边的行或列上，使用上面的算法，我会得到一个ArrayOutofBoundsException错误。

然后，我有一个巨大的if语句，用于判断这件作品是白的还是黑的。

我只是觉得有一个更简单的方法来优化这个…作为一个初学编程的人，我看不出来。有人能给我提点建议吗？

然后我必须再做一遍黑色的部分…我的代码看起来真的很邋遢，我觉得有一个更简单的方法来做这件事。有什么建议吗？

void makeMove(int typePiece, int fromRow, int fromCol, int toRow, int toCol) {
    board[toRow][toCol] = board[fromRow][fromCol];
    board[fromRow][fromCol] = EMPTY;        

    //CAPTURE
    if(typePiece == WHITE) {
        if(toRow==0) { //top row
            //check right
            if(toCol!=9 && board[toRow][toCol+1]==BLACK && 
                    (toCol==10 || board[toRow][toCol+2]==WHITE || board[toRow][toCol+2]==WHITEKING)) {
                board[toRow][toCol+1]=EMPTY;
            }
            //check left
            if(toCol!=1 && board[toRow][toCol-1]==BLACK && 
                    (toCol==0 || board[toRow][toCol-2]==WHITE || board[toRow][toCol-2]==WHITEKING)) {
                board[toRow][toCol-1]=EMPTY;
            }
            //check bottom
            if(board[toRow-1][toCol]==BLACK && (board[toRow-2][toCol]==WHITE || board[toRow-2][toCol]==WHITEKING)) {
                board[toRow-1][toCol]=EMPTY;
            }
        }
        else if(toRow == 10) { //bottom row
            //check right
            if(toCol!=9 && board[toRow][toCol+1]==BLACK && (toCol==10 || board[toRow][toCol+2]==WHITE || board[toRow][toCol+2]==WHITEKING)) {
                board[toRow][toCol+1]=EMPTY;
            }
            //check left
            if(toCol!=1 && board[toRow][toCol-1]==BLACK && (toCol==0 || board[toRow][toCol-2]==WHITE || board[toRow][toCol-2]==WHITEKING)) {
                board[toRow][toCol-1]=EMPTY;
            }
            //check top
            if(board[toRow+1][toCol]==BLACK && (board[toRow+2][toCol]==WHITE || board[toRow+2][toCol]==WHITEKING)) {
                board[toRow+1][toCol]=EMPTY;
            }
        }
        else if(toCol == 0) { //left column
            //check right
            if(board[toRow][toCol+1]==BLACK && (board[toRow][toCol+2]==WHITE || board[toRow][toCol+2]==WHITEKING)) {
                board[toRow][toCol+1]=EMPTY;
            }
            //check top
            if(toRow!=1 && board[toRow+1][toCol]==BLACK && (board[toRow+2][toCol]==WHITE || board[toRow+2][toCol]==WHITEKING)) {
                board[toRow+1][toCol]=EMPTY;
            }
            //check bottom
            if(toRow != 9 && board[toRow-1][toCol]==BLACK && (board[toRow-2][toCol]==WHITE || board[toRow-2][toCol]==WHITEKING)) {
                board[toRow-1][toCol]=EMPTY;
            }
        }
        else if(toCol == 10) { //right column
            //check left
            if(board[toRow][toCol-1]==BLACK && (toCol==0 || board[toRow][toCol-2]==WHITE || board[toRow][toCol-2]==WHITEKING)) {
                board[toRow][toCol-1]=EMPTY;
            //check top
            if(toRow!=1 && board[toRow+1][toCol]==BLACK && (board[toRow+2][toCol]==WHITE || board[toRow+2][toCol]==WHITEKING)) {
                board[toRow+1][toCol]=EMPTY;
            }
            //check bottom
            if(toRow != 9 && board[toRow-1][toCol]==BLACK && (board[toRow-2][toCol]==WHITE || board[toRow-2][toCol]==WHITEKING)) {
                board[toRow-1][toCol]=EMPTY;
            }
        }

Answer 1

考虑一个片段只能作为另一个片段运动的结果被捕获。假设你有一个白色的棋子移动到一个黑色棋子的附近（八个检查“我旁边有一个黑色棋子吗？”）；在这一点上，检查是否有一个白色片沿着同一条线从它对面（如果黑色片在第一个白色片的下面，检查在第一个黑色片的正下面的另一个白色片；如果黑色片在对角线上且向右，检查在黑色片的右边且向上的另一个白色片；等等）。此外，你需要检查每一个棋子移动a）是否有一个敌人的棋子在我周围，如果有b）它是镜像在另一边的另一个敌人的棋子？

尝试以下方法：

for all eight directions (up, down, left, right, and the four diagonals):
    is there an enemy piece in that direction adjacent to me?
    if so:
        is there a friendly piece in that direction that is also adjacent to that enemy piece?
        if so:
             remove that piece

而且

for all eight directions:
    is there an enemy piece in that direction adjacent to me?
    if so:
         is there another enemy piece in the opposite direction that is also adjacent to me?
    if so:
         remove me