用Java递归绘制图形

时间:2013-11-30 02:48:48

标签: java recursion graphics graphics2d

对于这个程序,我需要递归绘制一个'宝塔',这是一系列渐弱的矩形,在中央对齐,彼此叠加。我想我已经得到了实际数字背后的逻辑,但是我无法弄清楚如何将图形绘制为带有Graphics2D的矩形。我试图把它变成一个基本的形状绘图程序,但却找不到如何将它递归到它中。

这是我写到这一点的代码,没有考虑图形:

import java.awt.Rectangle;


public class PagodaDrawer
{

private int initialY; //Top of the bottom rectangle
private int initialHeight; //Height for the bottom rectangle
private double scale; //Amount to reduce each layer


public PagodaDrawer(int initialY, int initialHeight, double scaleFactor)
{
    this.initialY = initialY;
    this.initialHeight = initialHeight;
    scale = scaleFactor;
}

public void drawPagoda()
{
    drawLayer(0, initialY, 2 * initialHeight, initialHeight);
}

public void drawLayer(double x, double y, double width, double height)
{
    if(y < 0 || height < 5) //If off the top of the screen, or less than 5 tall
    {
        return;
    }
    drawLayer(x - (((1 - scale)* x) / 2), y + (y * scale), width * scale, height * scale );
    Rectangle r = new Rectangle((int)x, (int)y, (int)(2 * height), (int)height);
    //Draw r?
}
}

如何在框架中递归绘制图形的图层?

编辑:

对于任何感兴趣的人,这是最终的代码

import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Rectangle;
import java.util.ArrayList;

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


public class PagodaDrawer extends JPanel
{
private int initialX;
private int initialY; //Top of the bottom rectangle
private int initialHeight; //Height for the bottom rectangle
private double scale; //Amount to reduce each layer
private boolean isRenderable;
private ArrayList<Rectangle> recs;


public PagodaDrawer(int initialX, int initialY, int initialHeight, double scaleFactor)
{
    this.initialX = initialX;
    this.initialY = initialY;
    this.initialHeight = initialHeight;
    scale = scaleFactor;
    isRenderable = false;
    recs = new ArrayList<Rectangle>();
}

public void drawPagoda()
{
    drawLayer(initialX, initialY, 2 * initialHeight, initialHeight);
}

public void drawLayer(double x, double y, double width, double height)
{
    if(y < 0 || height < 5) //If off the top of the screen, or less than 5 tall
    {
        isRenderable = true;
        return;
    }
    drawLayer(x + .5 * (width - (width * scale)), y - (height * scale), width * scale, height * scale );
    Rectangle r = new Rectangle((int)x, (int)y, (int)(2 * height), (int)height);
    recs.add(r);
}

public void paintComponent(Graphics g)
{
    if(!isRenderable)
        return;
    super.paintComponent(g);
    Graphics2D g2 = (Graphics2D) g;
    for(int i = 0; i < recs.size(); i++)
    {
        g2.draw(recs.get(i));
        System.out.println(recs.get(i));
    }
}
}

加上这个JFrame:

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



public class DisplayComponent extends JFrame
{
private static final long serialVersionUID = -4279682826771265863L;
private static final int FRAME_WIDTH = 500;
private static final int FRAME_HEIGHT = 500;

private JPanel panel;
private PagodaDrawer p;

public DisplayComponent(int initialHeight, double scaleFactor)
{
    p = new PagodaDrawer(FRAME_WIDTH / 2, FRAME_HEIGHT, initialHeight, scaleFactor);
    panel = new JPanel();
    p.drawPagoda();
    add(p);

    pack();



    setSize(FRAME_WIDTH, FRAME_HEIGHT);
    setVisible(true);
}
}

2 个答案:

答案 0 :(得分:2)

不要使drawLayer()递归,而是编写一个递归createRectangle(),将每个新的Rectangle实例添加到List<Rectangle>。在paintComponent()的实施中渲染列表,图示为here

答案 1 :(得分:1)

在Java AWT和Swing中,您使用Graphics / Graphics2D方法绘制 示例:graphics.fillRect(x, y, w, h);

您应该从要绘制的组件中获取graphics(?:2d)对象,通常是主框架或某个组件。

在框架的paintComponent()内调用图纸应该可以正常工作:

How to use paintComponent in Java to paint multiple things, but rotate one?

这是Java6文档: http://docs.oracle.com/javase/6/docs/api/