我应该修改这个Sierpinski的三角形程序来计算三角形的数量。所以我试图在每次制作三角形时增加计数,但是,不知怎的,我的计数不会增加。
public class SierpinskiTriangle extends Applet
{
public int SeirpTri(Graphics g, int x1, int y1, int x2, int y2, int x3, int y3, int n, int count)
{
this.setBackground(new Color(0,0,0));
this.setSize(700, 500);
if ( n == 0 )
{
g.setColor(new Color(0, 255, 0));
g.drawLine(x1, y1, x2, y2); // if n = 0 draw the triangle
g.drawLine(x2, y2, x3, y3);
g.drawLine(x3, y3, x1, y1);
return 1;
}
int xa, ya, xb, yb, xc, yc; // make 3 new triangles by connecting the midpoints of
xa = (x1 + x2) / 2; //. the previous triangle
ya = (y1 + y2) / 2;
xb = (x1 + x3) / 2;
yb = (y1 + y3) / 2;
xc = (x2 + x3) / 2;
yc = (y2 + y3) / 2;
SeirpTri(g, x1, y1, xa, ya, xb, yb, n - 1, count++); // recursively call the function using the 3 triangles
SeirpTri(g, xa, ya, x2, y2, xc, yc, n - 1, count++);
SeirpTri(g, xb, yb, xc, yc, x3, y3, n - 1, count++);
return count;
}
public void paint(Graphics g)
{
int recursions = 3;
int count=1;
// call the recursive function sending in the number of recursions
SeirpTri(g, 319, 0, 0, 479, 639, 479, recursions, count);
// Counting triangles using math algorithm;
int count2 = 1;
if (recursions ==0) {
count2 =1;
}
else {
count2 = (int) Math.pow(3,(recursions-1)) * 3;
}
System.out.println("Correct answer is: " +count2);
System.out.println("Answer using recurvise is: " +count*3);
}
}
答案 0 :(得分:1)
您返回count
,但从不查看调用SeirpTri
的结果。
而不是:
SeirpTri(g, x1, y1, xa, ya, xb, yb, n - 1, count++); // recursively call the function using the 3 triangles
SeirpTri(g, xa, ya, x2, y2, xc, yc, n - 1, count++);
SeirpTri(g, xb, yb, xc, yc, x3, y3, n - 1, count++);
return count;
尝试类似:
return
SeirpTri(g, x1, y1, xa, ya, xb, yb, n - 1)
+ SeirpTri(g, xa, ya, x2, y2, xc, yc, n - 1)
+ SeirpTri(g, xb, yb, xc, yc, x3, y3, n - 1);
根本不需要count参数。每个SeirpTri
调用只需要知道它和它的“子”(在调用树上)创建的三角形。 “root”调用(在paint
中)将返回总计。
答案 1 :(得分:0)
每个参数都是用Java中的值传递的。这意味着对count
的任何更改只对方法是本地的,而不是更改从父方法传递给方法的count
对象。
你可以(和你的代码)通过返回count
参数来解决这个问题。您需要做的就是在父方法中设置count
。
替换每一行:
SeirpTri(...);
使用:
count = SeirpTri(...);