我需要采用两个相互递归方法的程序,并修改程序,使其包含单个递归方法。根据我的理解,我需要将递归调用按照调用顺序放置在单个方法中,以结合这两种递归方法。问题是有4个整数通过方法传递,第一个方法调用第二个方法两次,第二个方法调用第一个方法两次。在java中将两个相互递归的方法转换为单一的递归方法?
这是原来的代码:
public void drawHorizontal(Graphics graphics, double xMid, double yMid, double length)
{
// find left endpoint
double x1 = xMid - (length/2);
double y1 = yMid;
// find right endpoint
double x2 = xMid + (length/2);
double y2 = yMid;
if (length > 5)
{
// draw a line from (x1,y1) to (x2,y2)
graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2);
// draw a vertical line with left end of horizontal as midpoint of new line
drawVertical(graphics, x1, y1, (length));
// draw a vertical line with right endof horizontal as midpoint of new line
drawVertical(graphics, x2, y2, (length));
}
} // end drawHorizontal()
public void drawVertical(Graphics graphics, double xMid, double yMid, double length)
{
// find upper endpoint
double x1 = xMid;
double y1 = yMid - (length/2);
// right lower endpoint
double x2 = xMid;
double y2 = yMid + (length/2);
if (length > 5)
{
// draw a line from (x1,y1) to (x2,y2)
graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2);
// draw a 1/2 size horizontal line with top end of vertical as midpoint of new line
drawHorizontal(graphics, x1, y1, (length/2));
// draw a 1/2 horizontal line with bottom end of vertical as midpoint of new line
drawHorizontal(graphics, x2, y2, (length/2));
}
} // end drawVertical()
下面是我最近修改的代码。它的丑陋我知道,但我不知道如何独立调整x和y坐标。我试图通过创建更多变量来解决这个问题,但我不禁觉得我只是在做更多的工作。我能找到的最接近的堆栈问题是this。我从11号开始就已经在这里了,现在是4:15。向正确的方向微调将不胜感激,谢谢你的时间。
编辑*感谢您的快速回复,我很感激。我知道,以这种方式破解相互递归的方法似乎违反直觉,但我是编程和Java的新手,所以我正在探索解决问题的不同方法。这是我最终把它分解,并且运行良好。感谢您的时间。
修改后的代码:
public void Total(Graphics graphics, boolean type, double xMid, double yMid, double length) {
double x1;
double y1;
// find right endpoint
double x2;
double y2;
if (type == false) {
// find left endpoint
x1 = xMid - (length/2);
y1 = yMid;
// find right endpoint
x2 = xMid + (length/2);
y2 = yMid;
if (length > 5) {
// draw a line from (x1,y1) to (x2,y2)
graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2);
// draw a vertical line with left end of horizontal as midpoint of new line
Total(graphics, true, x1, y1, (length));
// draw a vertical line with right endof horizontal as midpoint of new line
Total(graphics, true, x2, y2, (length));
}
} else {
// find upper endpoint
x1 = xMid;
y1 = yMid - (length/2);
// right lower endpoint
x2 = xMid;
y2 = yMid + (length/2);
if (length > 5) {
// draw a line from (x1,y1) to (x2,y2)
graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2);
// draw a 1/2 size horizontal line with top end of vertical as midpoint of new line
Total(graphics, false, x1, y1, (length/2));
// draw a 1/2 horizontal line with bottom end of vertical as midpoint of new line
Total(graphics, false, x2, y2, (length/2));
}
}
}
为什么你需要*来做到这一点?我真的不明白它给了什么优势;你的方法非常明确。 –