使用旋转矩阵在Java中旋转矩形

Question

我在JPanel扩展类上绘制了一个矩形（一个Polygon对象）。

然后我尝试创建另一个Polygon对象，该对象应该是原始Polygon的90度旋转。

我使用了旋转矩阵。

由于某种原因，它不起作用。运行程序只显示原始矩形。

怎么了？感谢

public class DrawingPanel extends JPanel{

    public void paintComponent(Graphics g){

        super.paintComponent(g);
        Graphics2D g2d = (Graphics2D) g;

        Point p1,p2,p3,p4;
        p1 = new Point(50,50);
        p2 = new Point(200,50);
        p3 = new Point(200,100);
        p4 = new Point(50,100);

        int[] x = {(int) p1.getX(), (int) p2.getX(), (int)p3.getX(), (int) p4.getX()};
        int[] y = {(int) p1.getY(), (int) p2.getY(), (int)p3.getY(), (int) p4.getY()};

        Polygon poly = new Polygon(x, y, x.length);
        g2d.draw(poly);

        p1.setLocation(p1.getX() * Math.cos(Math.toRadians(90)) - p1.getY() * Math.sin(Math.toRadians(90)),
                p1.getX() * Math.sin(Math.toRadians(90)) + p1.getY() * Math.cos(Math.toRadians(90)));
        p2.setLocation(p2.getX() * Math.cos(Math.toRadians(90)) - p2.getY() * Math.sin(Math.toRadians(90)),
                p2.getX() * Math.sin(Math.toRadians(90)) + p2.getY() * Math.cos(Math.toRadians(90)));
        p3.setLocation(p3.getX() * Math.cos(Math.toRadians(90)) - p3.getY() * Math.sin(Math.toRadians(90)),
                p3.getX() * Math.sin(Math.toRadians(90)) + p3.getY() * Math.cos(Math.toRadians(90)));
        p4.setLocation(p4.getX() * Math.cos(Math.toRadians(90)) - p4.getY() * Math.sin(Math.toRadians(90)),
                p4.getX() * Math.sin(Math.toRadians(90)) + p4.getY() * Math.cos(Math.toRadians(90)));

        int[] x2 = {(int) p1.getX(), (int) p2.getX(), (int)p3.getX(), (int) p4.getX()};
        int[] y2 = {(int) p1.getY(), (int) p2.getY(), (int)p3.getY(), (int) p4.getY()};

        Polygon poly2 = new Polygon(x2, y2, x2.length);
        g2d.draw(poly2);

    }

}

Answer 1

我相信你正在将它从屏幕上旋转出来，所以我不会因此画出来。

您的代码将围绕左上角旋转90，导致位于屏幕左侧