Question

我有一个班级Grid，它管理所有地图功能。问题是，pacman地图逆时针旋转了90度。

看起来如何

The issue

它应该如何看待

The 'fixed' version

我通过在grid[x][y]内将grid[y][x]换成isWall()来获得'固定'版本（一种不整齐，不正确的方法）

以下是Grid类的完整代码;

package com.jackwilsdon.pacman.game;

import org.newdawn.slick.Graphics;

public class Grid {
    public static final int BLOCK_SIZE = 20;

    public int[][] grid = null;

    public Grid()
    {
        grid = new int[][]  {   {0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0},
                                {0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0},
                                {0,1,0,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,0,1,0},
                                {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
                                {0,1,0,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,0,1,0},
                                {0,1,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,0},
                                {0,1,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,0},
                                {0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,0},
                                {0,0,0,0,1,0,1,0,1,1,0,1,1,0,1,0,1,0,0,0,0},
                                {0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0},
                                {0,0,0,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,0,0,0},
                                {0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,0},
                                {0,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,0},
                                {0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0},
                                {0,1,0,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,0,1,0},
                                {0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0},
                                {0,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,0},
                                {0,1,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,0},
                                {0,1,0,1,1,1,1,1,1,0,1,0,1,1,1,1,1,1,0,1,0},
                                {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
                                {0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0}     };
    }

    public boolean isWall(int x, int y)
    {
        if (x >= 0 && x < grid.length && y >= 0 && y < grid[0].length)
        {
            return grid[y][x] == 1;
        }

        return true;
    }

    public void draw(Graphics g)
    {
        for (int cX = 0; cX < grid.length; cX++)
        {
            for (int cY = 0; cY < grid[cX].length; cY++)
            {
                if (this.isWall(cX, cY))
                {
                    g.fillRect(cX*Grid.BLOCK_SIZE, cY*Grid.BLOCK_SIZE, Grid.BLOCK_SIZE, Grid.BLOCK_SIZE);
                }
            }
        }
    }
}

我在代码中犯了一个愚蠢的错误吗？

我不想切换x和y，因为它不再是2d数组的正确格式。

Answer 1

我在代码中看到的问题是边界检查不正确。换句话说，这段代码：

if (x >= 0 && x < grid.length && y >= 0 && y < grid[0].length)
{
    return grid[y][x] == 1;
}

实际应该是

if (x >= 0 && x < grid[0].length && y >= 0 && y < grid.length)
{
    return grid[y][x] == 1;
}

您当前的代码仅适用于尺寸相等（地图为方形）。

draw函数中存在相同的错误。换句话说，它应该是

for (int cY = 0; cY < grid.length; cY++)
{
    for (int cX = 0; cX < grid[cY].length; cX++)
    {
        if (this.isWall(cX, cY))
        {
            g.fillRect(cX*Grid.BLOCK_SIZE, cY*Grid.BLOCK_SIZE, Grid.BLOCK_SIZE, Grid.BLOCK_SIZE);
        }
    }
}

在任何情况下，grid[y][x]都是正确的，而不是grid[x][y]，因为grid[i]指的是2D数组的索引i处的行，不是专栏。

Answer 2

这只是grid[i][j]引用j子数组中i个元素的事实的一个症状，实际上是j的位置第列和格式化网格中的i行。

由于您希望x代表该列，y代表该行，grid[y][x]是访问数组中某个位置的正确方法。

地图旋转90度？

2 个答案: