Java 在棋盘上放置12名骑士的算法，因此每个方块要么被占领，要么被攻击

我正在努力解决这个问题。我写了一些代码，一个试图找到这个问题的解决方案的递归，但是它需要很多时间来处理。我怎样才能使这种方法更有效

public static int placeKnights(Square current, int[][] board, int knightsPlaced) {
        ArrayList<Square> canAttackThisSquare = Square.canReachThisSquare(current); //Get a list of squares, that can attack our current square
        for (Square square : canAttackThisSquare) {     //Take each square from the list
            int[][] currBoard = Main.copyMatrix(board); //Copy the board, for next call
            Square curr = new Square(current.x, current.y);  //Copy the current square, for next call
            int knights = knightsPlaced + 1;  //Current knights placed (copying, so original knightsPlaced remain the same, to use when backtracking)
            ArrayList<Square> canBeAttackedFromSquare = Square.canReachThisSquare(square); //Get a list of squares, that can be attacked by the new knight
            for (Square sq : canBeAttackedFromSquare) {  //Put 1, to empty squares.
                if (currBoard[sq.x][sq.y] != 2) {
                    currBoard[sq.x][sq.y] = 1;
                }
            }
            // Get next square to attack
            while (board[current.x][current.y] != 0) {
                if (current.x == 7) {
                    if (current.y == 7) {
                        return knights;    //If the board is done then finish the recursion
                    } else {
                        current.x = 0;
                        current.y++;
                    }
                } else {
                    current.x++;
                }
            }
            if (knights == 12) {
                return 13;   //If the board is not done and we have already placed 12 knights, then end the recursion and return >12 value
            }
            if (placeKnights(curr, currBoard, knights) == 12) { //Else call the method on the next square, and get the result of recursion
                return 12;  //If the board got filled with 12 knights, then return 12
            }
            //Else backtrack and try next way
        }
        return 13;  //If it checks all the possible choices, it returns 13, so we backtrack.
    }

}

---

在为Board and Square编写类之后，我尝试了您的代码，但使用System.currentTimeMillis所花费的时间无法测量。所以我认为这部分代码没有问题。你能展示你的Square.canReachThisSquareSquare和你的Main.copyMatrixint[]]方法吗？有很多方法可以改进这段代码，即使不改变算法本身。只是一个想法：你可以用两个长值来表示你的棋盘：在第一个值中，如果有骑士，每个位都是1；在第二个值中，如果受到攻击，每个位都是1。这不仅使得复制非常便宜，而且可以通过简单的按位or实现骑士的放置。