Java需要太長時間才能從PNG圖像渲染圖塊

Question

我正在嘗試從PNG文件在（600/448）屏幕中渲染一個水平（1600/448）。 它嘗試渲染，但沒有成功，我認為這樣做會停止游戲循環。 我認為這是因為存在很多瓷磚，但我真的不知道該如何解決。

 public void loadLevel(BufferedImage image) {
    int width = image.getWidth();
    int height = image.getHeight();

    for (int xx = 0; xx < width; xx++) { //width = 1600
        for (int yy = 0; yy < height; yy++) { //height = 448;
            Color c = new Color(image.getRGB(xx,yy));
            int red = c.getRed();
            int green = c.getGreen();
            int blue = c.getBlue();

            if (red == 225 && blue == 225 & green == 225) {
                objectHandler.add(new GrassRight(xx * 32, yy * 32, ObjectId.GrassRight));
            } else if (red == 129 && blue == 250 & green == 212) {
                objectHandler.add(new Player(xx * 32, yy * 32, objectHandler, ObjectId.Player));
           } else if (red == 0 && blue == 0 && green == 0) {
               objectHandler.add(new Dirt(xx * 32, yy * 32, ObjectId.Dirt));

            } else if (red == 144 && blue == 144 & green == 213) {
                objectHandler.add(new Grass(xx * 32, yy * 32, ObjectId.Grass));

            } else if (red == 38 && blue == 56 & green == 50) {
                objectHandler.add(new GrassLeft(xx * 32, yy * 32, ObjectId.GrassLeft));

            } else if (red == 62 && blue == 35 & green == 39) {
                objectHandler.add(new SlideRight(xx * 32, yy * 32, ObjectId.SlideRight));

            } else if (red == 191 && blue == 12 & green == 54) {
                objectHandler.add(new SlideLeft(xx * 32, yy * 32, ObjectId.SlideLeft));

            } else if (red == 127 && blue == 194 & green == 87) {
                objectHandler.add(new DownSlideLeft(xx * 32, yy * 32, ObjectId.DownSlideLeft));

            } else if (red == 49 && blue == 146 & green == 27) {
                objectHandler.add(new DownSlideRight(xx * 32, yy * 32, ObjectId.DownSlideRight));
            }
        }
    }
}


public void init() {
    System.out.println("Loading...");
    LoadingHandler loadingHandler = new LoadingHandler();
    level = loadingHandler.loadImage("/Level1.png");
    objectHandler = new ObjectHandler();
    cameraHandler = new CameraHandler(0, 0);
    this.addKeyListener(new KeyHandler(objectHandler));
    loadLevel(level);
    System.out.println("Done!");
}

public void render() {
    BufferStrategy bs = this.getBufferStrategy();
    if (bs == null) {
        this.createBufferStrategy(3);
        return;
    }
    Graphics g = bs.getDrawGraphics();
    Graphics2D g2d = (Graphics2D) g;
    g.setColor(Color.BLACK);
    g.fillRect(0, 0, getWidth(), getHeight());
    g2d.translate(cameraHandler.getX(), cameraHandler.getY());
    System.out.println("Rendering!");
    objectHandler.render(g);
    System.out.println("Rendered!");
    g2d.translate(-cameraHandler.getX(), -cameraHandler.getY());
    g.dispose();
    bs.show();

}

public class ObjectHandler {
public LinkedList<ObjectManager> object = new LinkedList<>();

ObjectManager tempObject;

public void tick() {
    for (int i = 0; i < object.size(); i++) {
        tempObject = object.get(i);
        tempObject.tick(object);
    }
}

public void render(Graphics g) {
    for (int i = 0; i < object.size(); i++) {
        tempObject = object.get(i);
        tempObject.render(g);
    }
}

public void add(ObjectManager object) {
    this.object.add(object);

}

並給出以下輸出：

Loading...
Done!
Rendering!

PNG：

Answer 1

不必為每個像素創建一個對象（即716,800個對象），而是為每個矩形區域創建一個對象（15個對象）。

更新：

在ObjectManager類中，添加兩個int字段：width和height，然后呈現整個矩形而不是單個像素（或32x32正方形）。 您仍然可以使用PNG初始化對象，但是使用更智能的算法，該算法不會為每個像素重新創建新對象。