java - 如何从 drawSquare 方法中删除递归并获得完全相同的结果?

标签 java recursion coordinates computer-science square

我需要从 drawSquare 方法中删除递归。从该方法中删除递归后,我还有很多事情要做,但其余的我可以自己解决。我只需要一个无需递归即可完成完全相同事情的可行解决方案,我会找出其余部分。

下面是我制作 Square 类的方法:

import java.awt.Color;

public class Square {
    final int BLACK = Color.BLACK.getRGB();
    final int WHITE = Color.WHITE.getRGB();
    protected int center_x;
    protected int center_y;
    protected int side;
    protected int color;
    protected Square parentSquare;

    public Square(){
        this.center_x = 0;
        this.center_y = 0;
        this.side = 0;
        this.color = WHITE;
        this.parentSquare = null;
    }
    public Square(int center_x,int center_y,int side,int color){
        this.center_x = center_x;
        this.center_y = center_y;
        this.side = side;
        this.color = color;
        this.parentSquare = null;
    }
    public Square(int center_x,int center_y,int side,int color,Square parentSquare){
        this.center_x = center_x;
        this.center_y = center_y;
        this.side = side;
        this.color = color;
        this.parentSquare = parentSquare;
    }
    public void setX(int center_x){
        this.center_x = center_x;
    }
    public int getX(){
        return center_x;
    }
    public void setY(int center_y){
        this.center_x = center_y;
    }
    public int getY(){
        return center_y;
    }
    public void setSide(int side){
        this.side = side;
    }
    public int getSide(){
        return side;
    }
    public void setColor(int color){
        this.color = color;
    }
    public int getColor(){
        return color;
    }
    public void setParent(Square parentSquare){
        this.parentSquare = parentSquare;
    }
    public Square getParent(){
        return parentSquare;
    }
}

这是原始的 Tsquare.java,它产生一个正方形的分形,从每个正方形的 4 个角分支直到边 = 0:(完整的 TSquare.java 类修改为使用 Square 对象)

import java.awt.image.*;
import java.awt.Color;
import java.io.*;
import javax.imageio.*;
import java.util.*;
public class TSquare {
    static final int SIDE = 1000; // image is SIDE X SIDE
    static BufferedImage image = new BufferedImage(SIDE, SIDE, BufferedImage.TYPE_INT_RGB);
    static final int WHITE = Color.WHITE.getRGB();
    static final int BLACK = Color.BLACK.getRGB();
    static Scanner kbd = new Scanner(System.in);

    public static void main(String[] args) throws IOException{
        String fileOut = "helloSquares.png";
        System.out.print("Enter (x,y) coordinates with a space between: ");
        int x = kbd.nextInt();
        int y = kbd.nextInt();
        System.out.println(x+","+y);//TESTLINE TESTLINE TESTLINE TESTLINE
        // make image black
        for (int i = 0; i < SIDE; i++) {
            for (int j = 0; j < SIDE; j++) {
                image.setRGB(i, j, BLACK);
            }
        }
        Square square = new Square(SIDE/2,SIDE/2,SIDE/2,WHITE);
        drawSquare(square);



        // save image
        File outputfile = new File(fileOut);
        ImageIO.write(image, "jpg", outputfile);
    }

    private static void drawSquare(Square square){ // center of square is x,y length of side is s
        if (square.side <= 0){ // base case
            return;
        }else{
            // determine corners
            int left = square.center_x - (square.side/2);
            int top = square.center_y - (square.side/2);
            int right = square.center_x + (square.side/2);
            int bottom = square.center_y + (square.side/2);
            int newColor =square.color-100000;
            Square newSquareA = new Square(left,top,square.side/2,newColor);
            Square newSquareB = new Square(left,bottom,square.side/2,newColor);
            Square newSquareC = new Square(right,top,square.side/2,newColor);
            Square newSquareD = new Square(right,bottom,square.side/2,newColor);
            for (int i = left; i < right; i++){
                for (int j = top; j < bottom; j++){
                    image.setRGB(i, j, square.color);
                }
            }
            // recursively paint squares at the corners
            drawSquare(newSquareA);
            drawSquare(newSquareB);
            drawSquare(newSquareC);
            drawSquare(newSquareD);
        }

    }

}

我希望重现此代码的确切操作,只是减去递归,但我尝试的一切似乎都不起作用。我什至无法在原始黑色 Canvas 上显示一个白色方 block 。

最佳答案

如果我们想要在不影响速度的情况下提高可读性,我建议首先对 Square 添加一些内容:

public int half() {
    return side/2;
}
public int left() {
    return center_x - half();
}
public int top() {
    return center_y - half();
}
public int right() {
    return center_x + half();
}
public int bottom() {
    return center_y + half();
}

public void draw(BufferedImage image) {

    int left = left();
    int top = top();
    int right = right();
    int bottom = bottom();

    for (int i = left; i < right; i++){
        for (int j = top; j < bottom; j++){
            image.setRGB(i, j, color);
        }
    }
} //End Square

还将 I/O 移出以启用单元测试。

package com.stackoverflow.candied_orange;
import java.awt.image.*;
import java.awt.Color;
import java.io.*;
import javax.imageio.*;
import java.util.*;
public class FractalSquareIterative {

    public static void main(String[] args) throws IOException{

        final int SIDE = 1000; // image is SIDE X SIDE        
        BufferedImage image = new BufferedImage(SIDE,SIDE,BufferedImage.TYPE_INT_RGB);

        drawImage(SIDE, image);
        saveImage(image);
    }

    //Removed IO to enable unit testing
    protected static void drawImage(final int SIDE, BufferedImage image) {
        final int BLACK = Color.BLACK.getRGB();
        final int WHITE = Color.WHITE.getRGB();
        final int HALF = SIDE / 2;

        //Draw background on whole image
        new Square(HALF, HALF, SIDE, BLACK).draw(image);

        //Draw foreground starting with centered half sized square
        Square square = new Square(HALF, HALF, HALF, WHITE);
        drawFractal(square, image);
    }

现在 Square 正在处理所有正方形的东西,分形代码看起来更容易一些。

    private static void drawFractal(Square square, BufferedImage image){

        Queue<Square> squares = new LinkedList<>();
        squares.add(square);

        while (squares.size() > 0) {

            //Consume
            square = squares.remove();

            //Produce
            int half = square.half();
            if (half > 2) {

                int left = square.left();
                int top = square.top();
                int right = square.right();
                int bottom = square.bottom();
                int newColor = square.color - 100000;                

                squares.add(new Square(left, top, half, newColor));
                squares.add(new Square(left, bottom, half, newColor));
                squares.add(new Square(right, top, half, newColor));
                squares.add(new Square(right, bottom, half, newColor));
            }
            square.draw(image);
        }
    }


    protected static void saveImage(BufferedImage image) throws IOException {
        String fileOut = "helloSquares.png";        
        File outputfile = new File(fileOut);
        ImageIO.write(image, "jpg", outputfile);
    }
} //End FractalSquareIterative

可靠地比递归版本快,但在这个大小下并不显着。

如果您想看一下我的单元测试,您会发现它们 here .

关于java - 如何从 drawSquare 方法中删除递归并获得完全相同的结果?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57940778/

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