Java 在二维数组中搜索。

我有一种方法，可以确定给定行和列的棋盘上的正方形是否受到第1列到第1列中任何皇后的攻击。但是我在板[row-column+y][y]==QUEEN

  private boolean isUnderAttack(int row, int column)
  {
            for (int y=0; y<column; y++)
            {
             if (board[row][y] == QUEEN ||    // possible horizontal attack
             board[row-column+y][y] == QUEEN ||   // diagonal NW
             board[row+column-y][y] == QUEEN)     // diagonal SW

             return true;
            }

            return false;
  }

private boolean isUnderAttack（int行，int列）
{
对于（int y=0；yPay attation to:

row + column - y

此操作可以返回小于0的数字

你的班级应该是这样的：

import java.util.Arrays;

public class Board {

private final int[][] board;
private final int dimention;

public static void main(final String[] args) {
    final Board p = new Board(6);
    for (final int[] i : p.board) {
        System.out.println(Arrays.toString(i));
    }
    p.isUnderAttack(2, 3);
    System.out.println("\n\n");
    for (final int[] i : p.board) {
        System.out.println(Arrays.toString(i));
    }

}

public Board(final int dimention) {
    this.dimention = dimention;
    this.board = new int[dimention][dimention];
}

private void isUnderAttack(final int row, final int column) {

    for (int y = 0; y < this.dimention; y++) {
        this.board[y][row] = 1; // possible horizontal attack
        this.board[column][y] = 2; // possible vertical attack
    }
    int staringRow = column - row;
    int y = 0;
    do {
        this.board[staringRow][y] = 3; // diagonal SW
    } while ((++staringRow < this.dimention) && (++y < this.dimention));

    int staringCol = column + row;
    int x = 0;
    do {
        this.board[x][staringCol] = 4; // diagonal NW

    } while ((++x < this.dimention) && (--staringCol < this.dimention));
}

}

什么例外？ArrayIndexOutOfBounds？您应该提供更多信息，但问题是（很可能是）非常清楚。从异常中的索引，你应该能够找到它。在NW测试中，如果所讨论的正方形具有低行、低y和高列值，你将得到负索引。谢谢…不是我想要的。但是给了我一个想法
[0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0]



[0, 0, 1, 0, 0, 4]
[3, 0, 1, 0, 4, 0]
[0, 3, 1, 4, 0, 0]
[2, 2, 4, 2, 2, 2]
[0, 4, 1, 3, 0, 0]
[4, 0, 1, 0, 3, 0]