斐波那契迹线

托马斯·斯科特

我只想事先知道这是一项家庭作业。我已经以许多不同的方式尝试了该问题,以至于我对为什么我没有得到期望的输出一无所知。

问题

编写一个程序,该程序将跟踪如何递归生成斐波那契数(对于任何N),并以以下方式显示跟踪:

范例(N = 4):

Entering level 0
Entering level 2
Entering level 4
Exiting level 4
Entering level 3
Exiting level 3
Exiting level 2
Entering level 1
Entering level 3
Exiting level 3
Entering level 2
Entering level 4
Exiting level 4
Entering level 3
Exiting level 3
Exiting level 2
Exiting level 1
Exiting level 0

我的主:

public class A5main {

public static void main(String[] args) {
    //n holds user input
    //level is the current level of the tree
    //fibonacci is a5class object
    int n;
    int level=0;
    a5class fibonacci= new a5class();
    Scanner keyboard = new Scanner(System.in);

    //Ask user for input

   System.out.println("Enter a number up to which Fibonacci series to print: ");
   n = keyboard.nextInt();
   System.out.println("Fibonacci trace of: " + n);

    //Pass input to fibonacci.trace method with arguments n, level.
   fibonacci.trace(n,level);

}
}

我的课:

 package a5main;
 public class a5class {
    int fibWork;
   public a5class()     
{ 
}
public int trace(int t, int level)
{
    //Accepts t and level as an argument.
    //Lets uer know what level they are entering 
    System.out.println("Now entering level " + level);
    //If t<=1 just return the value
        if (t<=1)
        {
            System.out.println ("\tNow exiting level " + level);
            return t;

        }
        //Else use recurssion to figure out the fibonacci sequence
        //and determine what level you are on.
        else
        {
            fibWork = trace(t-1, level+1) + trace(t-2, level+1);
            System.out.println ("\tNow exiting level " + level);
            return t; 
        }
}
}

(我把\ t放到那里,这样我现在就可以看到它的退出位置更容易了)

我的输出:

Enter a number up to which Fibonacci series to print: 
4
Fibonacci trace of: 4
Now entering level 0
Now entering level 1
Now entering level 2
Now entering level 3
     Now exiting level 3
Now entering level 3
     Now exiting level 3
     Now exiting level 2
Now entering level 2
     Now exiting level 2
     Now exiting level 1
Now entering level 1
Now entering level 2
     Now exiting level 2
Now entering level 2
     Now exiting level 2
     Now exiting level 1
     Now exiting level 0

我还尝试不将'level传递给该方法,并在创建对象时在我的公共a5class中将其等于0,但到目前为止还算不上成功。

虽然我的离我可能还不算太远,但我注意到它退出了3级,然后在下一个顺序中重新输入了它,这在逻辑上似乎也不正确。

感谢您的帮助或指导。即使只是指向正确方向的指针也值得赞赏。我认真对待我的编程,想真正理解它。我不想“伪造”自己的方式并没有价值。

谢谢!

乔尼

从预期的输出中,您将看到程序希望直接从0级进入2级,并从2级进入4级。只有在第一次递归调用中将2级添加到该级时,才会发生这种情况:

fibWork = trace(t-2, level+2) + trace(t-1, level+1);

本文收集自互联网,转载请注明来源。

如有侵权,请联系[email protected] 删除。

编辑于
0

我来说两句

0条评论
登录后参与评论

相关文章