今天来写k阶科赫雪花的递归实现,(K值需要你手动输入)至于科赫雪花是什么请大家自行百度。
首先来思考这个程序怎么写,当 count = 0 时就应该是一个三角形,这三个点是你一开始就确定的,以后的改变都依据这三个点发展的。当不是0的时候就需要计算相对于这个三角形的9个点,分别是每条边上的两个点,和它对应的三角形第三个顶点。
首先在JFrame中添加一个panel,我们需要在这个panel上画图。
大家再来看这个图片,这张图介绍了通过两个点来计算其他三个点的过程。
现在开始在panel中画图:
static class showpanel extends JPanel{ int number = 0; public void setNumber(int number) { this.number = number; repaint(); } public void paintComponent(Graphics g) { super.paintComponent(g);//画一个简单的panel int side =(int)(Math.min((int)getWidth(),(int)getHeight())*0.8); int high =(int)(side*Math.cos(Math.toRadians(30))); Point p1 = new Point(getWidth() / 2, 10); Point p2 = new Point(getWidth() / 2 - side / 2, 10 + high); Point p3 = new Point(getWidth() / 2 + side / 2, 10 + high); playKochSnowFlake(g, number, p1, p2); playKochSnowFlake(g, number, p2, p3); playKochSnowFlake(g, number, p3, p1); }
现在开始写递归函数。
public static void playKochSnowFlake(Graphics g,int number,Point p1,Point p2) { if(number == 0){ g.drawLine(p1.x, p1.y,p2.x, p2.y); } else{ int deltaX = p2.x - p1.x; int deltaY = p2.y - p1.y; Point x = new Point(p1.x + deltaX / 3, p1.y + deltaY / 3); Point y = new Point(p1.x + deltaX * 2 / 3, p1.y + deltaY * 2 / 3); Point z = new Point( (int)((p1.x + p2.x) / 2 + Math.sin(Math.toRadians(60)) * (p1.y - p2.y) / 3), (int)((p1.y + p2.y) / 2 + Math.sin(Math.toRadians(60)) * (p2.x - p1.x) / 3)); playKochSnowFlake(g, number - 1, p1, x); playKochSnowFlake(g, number - 1, x, z); playKochSnowFlake(g, number - 1, z, y); playKochSnowFlake(g, number - 1, y, p2); } }
然后在主面板中加入一个JTextField jta 它输入的数据要传入到number中。所以为其添加一个监听器。 已有数据输入就调用其中的setNumber()函数设置number变量。
jta.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent arg0) { spl.setNumber(Integer.parseInt(jta.getText())); } });
所以总体已经完成了,剩下的就是简答的窗体设置。
下面贴一个完整的java代码:
import java.awt.BorderLayout; import java.awt.FlowLayout; import java.awt.Graphics; import java.awt.Point; import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import javax.swing.JFrame; import javax.swing.JLabel; import javax.swing.JPanel; import javax.swing.JTextField; public class SnowFlake extends JFrame { private JTextField jta = new JTextField(5); private showpanel spl = new showpanel(); static class showpanel extends JPanel{ int number = 0; public void setNumber(int number) { this.number = number; repaint(); } public void paintComponent(Graphics g) { super.paintComponent(g);//画一个简单的panel int side =(int)(Math.min((int)getWidth(),(int)getHeight())*0.8); int high =(int)(side*Math.cos(Math.toRadians(30))); Point p1 = new Point(getWidth() / 2, 10); Point p2 = new Point(getWidth() / 2 - side / 2, 10 + high); Point p3 = new Point(getWidth() / 2 + side / 2, 10 + high); playKochSnowFlake(g, number, p1, p2); playKochSnowFlake(g, number, p2, p3); playKochSnowFlake(g, number, p3, p1); } public static void playKochSnowFlake(Graphics g,int number,Point p1,Point p2) { if(number == 0){ g.drawLine(p1.x, p1.y,p2.x, p2.y); } else{ int deltaX = p2.x - p1.x; int deltaY = p2.y - p1.y; Point x = new Point(p1.x + deltaX / 3, p1.y + deltaY / 3); Point y = new Point(p1.x + deltaX * 2 / 3, p1.y + deltaY * 2 / 3); Point z = new Point( (int)((p1.x + p2.x) / 2 + Math.sin(Math.toRadians(60)) * (p1.y - p2.y) / 3), (int)((p1.y + p2.y) / 2 + Math.sin(Math.toRadians(60)) * (p2.x - p1.x) / 3)); playKochSnowFlake(g, number - 1, p1, x); playKochSnowFlake(g, number - 1, x, z); playKochSnowFlake(g, number - 1, z, y); playKochSnowFlake(g, number - 1, y, p2); } } } public SnowFlake() { JPanel panel = new JPanel(); panel.setLayout(new FlowLayout()); panel.add(new JLabel("Please input the number")); panel.add(jta); add(spl,BorderLayout.CENTER); add(panel,BorderLayout.SOUTH); jta.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent arg0) { spl.setNumber(Integer.parseInt(jta.getText())); } }); } public static void main(String args[]) { SnowFlake snowFlake = new SnowFlake(); snowFlake.setSize(300, 300); snowFlake.setTitle("SnowFlake"); snowFlake.setLocationRelativeTo(null); snowFlake.setVisible(true); } }
效果图:
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持鸟哥教程(niaoge.com)。
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