如何将代码转换为递归?
how to convert code into recursion?
import java.awt.Graphics;
import javax.swing.JComponent;
import javax.swing.JFrame;
public class DrawIt
{
public static void main(String[] args)
{
JFrame frame = new JFrame();
final int width = 400;
final int height = 400;
frame.setSize(width,height);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
JComponent component = new JComponent()
{
public void paintComponent(Graphics graph){
draw(graph);
}
};
frame.add(component);
frame.setVisible(true);
}
public static void draw(Graphics g)
{
int x1=100;
int y1 = 100;
int length = 10;
for(int i=0;i<=10;i++)
{
int x2 = x1 + length;
int y2 = y1;
g.drawLine(x1,y1,x2,y2);
x1=x2;
y1=y2;
y2=y1-length;
g.drawLine(x1,y1,x2,y2);
x1=x2;
y1=y2;
length+=10;
x2=x1-length;
g.drawLine(x1, y1, x2, y2);
x1=x2;
y1=y2;
y2=y1+length;
g.drawLine(x1, y1, x2, y2);
x1=x2;
y1=y2;
length+=10;
}
}
}
如何将其转换为递归?这给了我矩形螺旋但不是递归方式。帮助。如果这样我使用 4 个变量,递归,它会怎样?
public class DrawIt {
public static void main(String[] args) {
JFrame frame = new JFrame();
final int width = 400;
final int height = 400;
frame.setSize(width, height);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
JComponent component = new JComponent() {
public void paintComponent(Graphics graph) {
draw(graph);
}
};
frame.add(component);
frame.setVisible(true);
}
public static void drawpuzzle(Graphics g, int x1, int y1, int length, int count) {
if (count> 0) {
int x2 = x1 + length;
int y2 = y1;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
y2 = y1 - length;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
length += 10;
x2 = x1 - length;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
y2 = y1 + length;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
length += 10;
drawpuzzle(g, x1, y1, length, count - 1);
}
}
public static void draw(Graphics g) {
int x1 = 100;
int y1 = 100;
int length = 10;
int count = 10;
drawpuzzle(g, x1, y1, length, count);
}
}
import java.awt.Graphics;
import javax.swing.JComponent;
import javax.swing.JFrame;
public class DrawIt
{
public static void main(String[] args)
{
JFrame frame = new JFrame();
final int width = 400;
final int height = 400;
frame.setSize(width,height);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
JComponent component = new JComponent()
{
public void paintComponent(Graphics graph){
draw(graph);
}
};
frame.add(component);
frame.setVisible(true);
}
public static void draw(Graphics g)
{
int x1=100;
int y1 = 100;
int length = 10;
for(int i=0;i<=10;i++)
{
int x2 = x1 + length;
int y2 = y1;
g.drawLine(x1,y1,x2,y2);
x1=x2;
y1=y2;
y2=y1-length;
g.drawLine(x1,y1,x2,y2);
x1=x2;
y1=y2;
length+=10;
x2=x1-length;
g.drawLine(x1, y1, x2, y2);
x1=x2;
y1=y2;
y2=y1+length;
g.drawLine(x1, y1, x2, y2);
x1=x2;
y1=y2;
length+=10;
}
}
}
如何将其转换为递归?这给了我矩形螺旋但不是递归方式。帮助。如果这样我使用 4 个变量,递归,它会怎样?
public class DrawIt {
public static void main(String[] args) {
JFrame frame = new JFrame();
final int width = 400;
final int height = 400;
frame.setSize(width, height);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
JComponent component = new JComponent() {
public void paintComponent(Graphics graph) {
draw(graph);
}
};
frame.add(component);
frame.setVisible(true);
}
public static void drawpuzzle(Graphics g, int x1, int y1, int length, int count) {
if (count> 0) {
int x2 = x1 + length;
int y2 = y1;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
y2 = y1 - length;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
length += 10;
x2 = x1 - length;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
y2 = y1 + length;
g.drawLine(x1, y1, x2, y2);
x1 = x2;
y1 = y2;
length += 10;
drawpuzzle(g, x1, y1, length, count - 1);
}
}
public static void draw(Graphics g) {
int x1 = 100;
int y1 = 100;
int length = 10;
int count = 10;
drawpuzzle(g, x1, y1, length, count);
}
}