编写程序实现谢尔宾斯基三角形

废话不说，只看代码！嘻嘻。。。。

import java.awt.Color;
import java.awt.Dimension;
import java.awt.Graphics;
import java.awt.Toolkit;
import java.awt.event.MouseAdapter;
import java.awt.event.MouseEvent;
import javax.swing.JFrame;
import javax.swing.JPanel;
/*** 画出一个谢尔宾斯基三角形* @author LONG**/
public class SanJiao extends JFrame {private static final long serialVersionUID = 1L;Dimension di = null;Graphics gr = null;JPanel jp = null;/*** 主函数，程序的入口* @param args*/public static void main(String[] args){SanJiao sj = new SanJiao();sj.showFrame();}/*** 设置用来画出图形的面板*/public void showFrame(){this.setTitle("画图");Toolkit tl = Toolkit.getDefaultToolkit();di = tl.getScreenSize();this.setSize(di.width,di.height);this.setDefaultCloseOperation(3);jp = new JPanel();jp.setPreferredSize(new Dimension(600,400));this.add(jp);this.setResizable(false);jp.setBackground(Color.BLACK);this.setVisible(true);gr = jp.getGraphics();jp.addMouseListener(new MouseAdapter(){public void mousePressed(MouseEvent e){}public void mouseReleased(MouseEvent e){draw(e);}});/*** 画出一个谢尔宾斯基三角形的准备工作，设置好显示的位置*/}public void draw(MouseEvent e){int x1 = di.width/4;int y1 = di.height*4/5;int x2 = di.width*3/4;int y2 = di.height*4/5;doSomething(x1,y1,x2,y2);}/*** 最终用来递归的函数，来递归画出三角形* @param x1	传入的左x坐标* @param y1	传入的左y坐标* @param x2	传入的右x坐标* @param y2	传入的右y坐标*/public void doSomething(int x1,int y1,int x2,int y2){//控制线程，可以更清楚的看见递归执行的过程try {Thread.sleep(10);} catch (InterruptedException e) {e.printStackTrace();}//设置控制条件为，最终画的三角形两底边的x坐标值之差大于10个像素if(Math.max(x2, x1) - Math.min(x2, x1) > 10){int x3 = (x1 + x2)/2;		//用来计算顶点坐标x值int y3 = (int)((y1 + y2)/2 - (Math.max(x2, x1) - Math.min(x2, x1))*Math.sqrt(3)/2);	//用来计算定点坐标的y值//为了绚丽一点，设置画笔颜色为随机色gr.setColor(new Color((int)(Math.random()*255),(int)(Math.random()*255),(int)(Math.random()*255)));//将三条边连接起来gr.drawLine(x1,y1,x2,y2);gr.drawLine(x1,y1,x3,y3);gr.drawLine(x2,y2,x3,y3);//得到三条边的中点坐标int c_x1 = (x1 + x2)/2;int c_y1 = (y1 + y2)/2;int c_x2 = (x1 + x3)/2;int c_y2 = (y1 + y3)/2;int c_x3 = (x2 + x3)/2;int c_y3 = (y2 + y3)/2;//在此处，递归时，先把一个画完，然后才会执行另外的一个，所以关键步骤是要明确终止递归的条件gr.drawLine(c_x1, c_y1, c_x2, c_y2);gr.drawLine(c_x1, c_y1, c_x3, c_y3);gr.drawLine(c_x2, c_y2, c_x3, c_y3);doSomething(x1,y1,c_x1,c_y1);doSomething(c_x1,c_y1,x2,y2);doSomething(c_x2,c_y2,c_x3,c_y3);}}}