使用javaswing，我有一条水平线（作为地平线）。我想画矩形，堆叠，一个在另一个上面，最大宽度的矩形在底部，最小宽度的矩形在顶部。当用户重新调整屏幕大小时，我还希望矩形保持在地平线上。这是我到目前为止的计算结果：

package testing;
import java.awt.BorderLayout;
import java.awt.Color;
import java.awt.Dimension;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Rectangle;
import java.awt.RenderingHints;
import java.util.ArrayList;
import java.util.List;

import javax.swing.JComponent;
import javax.swing.JFrame;
import javax.swing.JPanel;

public class Test extends JFrame {
    private static final int FRAME_WIDTH = 600;
    private static final int FRAME_HEIGHT = 500;

    public static void main(String[] args) {
        new Test();
    }

    protected Test() {
        this.setTitle("Tower of Hanoi");
        this.setSize(FRAME_WIDTH, FRAME_HEIGHT);
        this.setPreferredSize(new Dimension(FRAME_WIDTH, FRAME_HEIGHT));
        this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        this.setMinimumSize(new Dimension(FRAME_WIDTH, FRAME_HEIGHT));
        this.setMaximumSize(new Dimension(FRAME_WIDTH, FRAME_HEIGHT));
        this.setSize(new Dimension(FRAME_WIDTH, FRAME_HEIGHT));
        this.setBackground(Color.BLUE);
        this.add(new PanelTest(), BorderLayout.CENTER);
        this.setVisible(true);
    }

    public class PanelTest extends JPanel {

        private static final int FRAME_WIDTH = 600;
        private static final int FRAME_HEIGHT = 500;

        protected PanelTest() {
            setLayout(new BorderLayout());
            this.setSize(FRAME_WIDTH, FRAME_HEIGHT);
            this.setMinimumSize(new Dimension(FRAME_WIDTH, FRAME_HEIGHT));
            this.setMaximumSize(new Dimension(FRAME_WIDTH, FRAME_HEIGHT));
            this.setBackground(Color.CYAN);
            this.add(new ComponentTest(), BorderLayout.CENTER);
            revalidate();
            repaint();
            this.setVisible(true); //probably not necessary

        }
    }

    public class ComponentTest extends JComponent {
        private List<Rectangle> rects; 

        public ComponentTest() {
            rects = new ArrayList<>(); 
            rects.add(new Rectangle(30, 30, 30, 30));
            rects.add(new Rectangle(30, 30, 30, 30));
            rects.add(new Rectangle(30, 30, 30, 30));
            rects.add(new Rectangle(30, 30, 30, 30));
        }

        @Override
        protected void paintComponent(Graphics g) {
            super.paintComponent(g);

            Graphics2D graph2 = (Graphics2D) g;
            graph2.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);

            paint(g);
        }

        public void paint(Graphics g) {
            // Draw the platform
            g.drawLine(100, getHeight() - 100, getWidth() - 100, getHeight() - 100);

            for(int i = 0; i < rects.size(); i++) {
                rects.get(i).setBounds(getWidth()/2,
                        (30 * i + 1),
                        30 * i + 1,
                        30
                        );
                g.drawRect(rects.get(i).x, rects.get(i).y, rects.get(i).width, rects.get(i).height);
            }
        }
    }
}

这将使矩形堆叠起来，但浮动在屏幕顶部附近：

我怎样才能让它粘在黑色的地平线上？换句话说，如何让最大的矩形停留在黑线之上？此外，当屏幕被拖动到一个更大的尺寸-我如何增长所有的矩形高度（平等）和它仍然坚持在黑色的地平线？
还要注意的是，被绘制的矩形的数量是由用户决定的（在本例中是5，但最多可以是9）。
执行此操作也无法按预期工作：

for(int i = 0; i < rects.size(); i++) {
        rects.get(i).setBounds(getWidth()/2,
                (30 * i + 1),
                getHeight() - 100 - (30 * (rects.size() - (i + 1))),
                30
                );
        g.drawRect(rects.get(i).x, rects.get(i).y, rects.get(i).width, rects.get(i).height);
    }