改善繪製畢達哥拉斯樹

Question

我已經編寫繪製畢達哥拉斯樹分形的程序。任何人都可以看到改進它的任何方法嗎？現在它是89 LOC。

import java.awt.*; 
import java.util.Scanner; 
import javax.swing.*; 

public class Main extends JFrame {; 

    public Main(int n) { 
     setSize(900, 900); 
     setTitle("Pythagoras tree"); 
     add(new Draw(n)); 
     setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); 
     setVisible(true); 
    } 

    public static void main(String[] args) { 
     Scanner sc = new Scanner(System.in); 
     System.out.print("Give amount of steps: "); 
     new Main(sc.nextInt()); 
    } 
} 

class Draw extends JComponent { 
    private int height = 800; 
    private int width = 800; 
    private int steps; 

    public Draw(int n) { 
     steps = n; 

     Dimension d = new Dimension(width, height); 
     setMinimumSize(d); 
     setPreferredSize(d); 
     setMaximumSize(d); 
    } 

    @Override 
    public void paintComponent(Graphics g) { 
     super.paintComponent(g); 
     g.setColor(Color.white); 
     g.fillRect(0, 0, width, height); 
     g.setColor(Color.black); 

     int x1, x2, x3, y1, y2, y3; 
     int base = width/7; 

     x1 = (width/2)-(base/2); 
     x2 = (width/2)+(base/2); 
     x3 = width/2; 
     y1 = (height-(height/15))-base; 
     y2 = height-(height/15); 
     y3 = (height-(height/15))-(base+(base/2)); 

     g.drawPolygon(new int[]{x1, x1, x2, x2, x1}, new int[]{y1, y2, y2, y1, y1}, 5); 

     int n1 = steps; 
     if(--n1 > 0){ 
      g.drawPolygon(new int[] {x1, x3, x2}, new int[] {y1, y3, y1}, 3); 
      paintMore(n1, g, x1, x3, x2, y1, y3, y1); 
      paintMore(n1, g, x2, x3, x1, y1, y3, y1); 
     } 
    } 

    public void paintMore(int n1, Graphics g, double x1_1, double x2_1, double x3_1, double y1_1, double y2_1, double y3_1){ 
     int x1, x2, x3, y1, y2, y3; 

     x1 = (int)(x1_1 + (x2_1-x3_1)); 
     x2 = (int)(x2_1 + (x2_1-x3_1)); 
     x3 = (int)(((x2_1 + (x2_1-x3_1)) + ((x2_1-x3_1)/2)) + ((x1_1-x2_1)/2)); 
     y1 = (int)(y1_1 + (y2_1-y3_1)); 
     y2 = (int)(y2_1 + (y2_1-y3_1)); 
     y3 = (int)(((y1_1 + (y2_1-y3_1)) + ((y2_1-y1_1)/2)) + ((y2_1-y3_1)/2)); 

     g.setColor(Color.green); 
     g.drawPolygon(new int[] {x1, x2, (int)x2_1, x1}, new int[] {y1, y2, (int)y2_1, y1}, 4); 
     g.drawLine((int)x1, (int)y1, (int)x1_1, (int)y1_1); 
     g.drawLine((int)x2_1, (int)y2_1, (int)x2, (int)y2); 
     g.drawLine((int)x1, (int)y1, (int)x2, (int)y2); 

     if(--n1 > 0){ 
      g.drawLine((int)x1, (int)y1, (int)x3, (int)y3); 
      g.drawLine((int)x2, (int)y2, (int)x3, (int)y3); 
      paintMore(n1, g, x1, x3, x2, y1, y3, y2); 
      paintMore(n1, g, x2, x3, x1, y2, y3, y1); 
     } 
    } 
} 

Answer 1

• 使用drawPolygon，而不是乘drawLine
• 刪除不使用的方法pow，
• 清理進口
• 刪除不必要的變量h和w
• 把--n > 0成一條線
• 做一些重新格式化

和你去那裏，90線（註釋仍然計算在內）：

import java.awt.*; 
import java.util.Scanner; 
import javax.swing.*; 

public class Main extends JFrame {; 

    public Main(int n) { 
     setSize(900, 900); 
     setTitle("Pythagoras tree"); 
     add(new Draw(n)); 
     setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); 
     setVisible(true); 
    } 

    public static void main(String[] args) { 
     Scanner sc = new Scanner(System.in); 
     System.out.print("Give amount of steps: "); 
     new Main(sc.nextInt()); 
    } 
} 

class Draw extends JComponent { 
    private int height = 800; 
    private int width = 800; 
    private int steps; 

    public Draw(int n) { 
     steps = n; 

     Dimension d = new Dimension(width, height); 
     setMinimumSize(d); 
     setPreferredSize(d); 
     setMaximumSize(d); 
    } 

    @Override 
    public void paintComponent(Graphics g) { 
     super.paintComponent(g); 
     g.setColor(Color.white); 
     g.fillRect(0, 0, width, height); 
     g.setColor(Color.black); 

     int x1, x2, x3, y1, y2, y3; 
     int base = width/7; 

     x1 = (width/2)-(base/2); 
     x2 = (width/2)+(base/2); 
     x3 = width/2; 
     y1 = (height-(height/15))-base; 
     y2 = height-(height/15); 
     y3 = (height-(height/15))-(base+(base/2)); 

     //paint 
     g.drawPolygon(new int[]{x1, x1, x2, x2, x1}, new int[]{y1, y2, y2, y1, y1}, 5); 

     int n1 = steps; 
     if(--n1 > 0){ 
      g.drawPolygon(new int[] {x1, x3, x2}, new int[] {y1, y3, y1}, 3); 
      paintMore(n1, g, x1, x3, x2, y1, y3, y1); 
      paintMore(n1, g, x2, x3, x1, y1, y3, y1); 
     } 
    } 

    public void paintMore(int n1, Graphics g, double x1_1, double x2_1, double x3_1, double y1_1, double y2_1, double y3_1){ 
     double x1, x2, x3, y1, y2, y3; 
     //counting 
     x1 = x1_1 + (x2_1-x3_1); 
     x2 = x2_1 + (x2_1-x3_1); 
     x3 = ((x2_1 + (x2_1-x3_1)) + ((x2_1-x3_1)/2)) + ((x1_1-x2_1)/2); 
     y1 = y1_1 + (y2_1-y3_1); 
     y2 = y2_1 + (y2_1-y3_1); 
     y3 = ((y1_1 + (y2_1-y3_1)) + ((y2_1-y1_1)/2)) + ((y2_1-y3_1)/2); 

     //paint 
     g.setColor(Color.green); 
     g.drawPolygon(new int[] {(int)x1, (int)x2, (int)x2_1, (int)x1}, 
         new int[] {(int)y1, (int)y2, (int)y2_1, (int)y1}, 4); 

     g.drawLine((int)x1, (int)y1, (int)x1_1, (int)y1_1); 
     g.drawLine((int)x2_1, (int)y2_1, (int)x2, (int)y2); 
     g.drawLine((int)x1, (int)y1, (int)x2, (int)y2); 

     if(--n1 > 0){ 
      g.drawLine((int)x1, (int)y1, (int)x3, (int)y3); 
      g.drawLine((int)x2, (int)y2, (int)x3, (int)y3); 
      paintMore(n1, g, x1, x3, x2, y1, y3, y2); 
      paintMore(n1, g, x2, x3, x1, y2, y3, y1); 
     } 
    } 
} 

Answer 2

取代：

private int pow(int n){ 

    int pow = 2; 
    for(int i = 1; i < n; i++){ 
     if(n==0){ 
      pow = 1; 
     } 
     pow = pow*2; 
    } 
    return pow; 
} 

與

private int pow2(int n) { //Better name, to avoid confusion 
    return 1 << n; 
} 

雖然我不能看到你使用此功能。

Answer 3

重寫它25 lines of F#？ ;-)

Answer 4

有許多縮短代碼的方法（Eric和Mihir的建議是一個好的開始），但我會專注於清晰度。像x1和y1這樣的變量尖叫一個Point結構來將這些對保存在一起。然後，您可以將這些點傳遞給一個數組。