递归练习

Question

所以我遇到了以下问题： 我们有一只兔子站在一条线上，编号为1,2，......奇怪的兔子（1,3，......）有正常的2只耳朵。我们所说的平均兔子（2,4，...）有3只耳朵，因为它们每只都有一个凸起的脚。递归地返回兔子行1,2，... n中的“耳朵”数量（没有循环或乘法）。

这是我的源代码。我正在使用Java

import java.util.Scanner;

public class BunnyEars
{
    public static int CountEars(int i, int e)
    {
        if(i == 0)
        {
            return e;
        }
        else if(i > 0)
        {
            if(i%2==0)
            {
                e = e + 2;
                i = i - 1;
                CountEars(i,e);
            }
            else
            {
                e = e + 3;
                i = i - 1;
                CountEars(i,e);
            }       
        }
        return e;
    }

public static void main(String []args)
{
    Scanner scan = new Scanner(System.in);

    int b;
    int result;

    System.out.println("Bunny Ears, even has 2 ears, odd has 3 ears");
    System.out.println("Please enter a value: ");
    b = scan.nextInt();

    result = CountEars(b,0);

    System.out.println("Number of ears are: " + result);
}

如果输入5，则输出应为12，但输出为3。 所以我认为

中的CountEars（i，e）方法

enter code here

if(i%2==0)
        {
            e = e + 2;
            i = i - 1;
            CountEars(i,e);
        }
        else
        {
            e = e + 3;
            i = i - 1;
            CountEars(i,e);
        }   

未执行。我似乎无法找到我的错误。任何人？

Answer 1

在你的递归方法中，你并没有总计来自行中其他兔子的返回值。此外，您只需要一个参数 - 正在考虑哪个兔子。

/*
 * Return the total number of ears in a line of length n
 */
public static int countEars(int n) {
   if (n < 1) {
      return 0;   // base case: no bunnies, no ears
   }
   // Below we make the "recursive leap of faith": if this
   // function works, we can get the answer for a line of n-1
   // bunnies and add the current bunny's ears to it to yield
   // the correct answer for the current value of n.
   // Note that the argument to the recursive call must
   // converge towards the base case for this to work.
   if (n % 2 == 1) {
      return 2 + countEars(n-1);
   }
   return 3 + countEars(n-1);
}

使用result = countEars(5);调用此内容。

请注意，我已将您的方法重命名为遵循Java约定。

Answer 2

试试这段代码

import java.util.Scanner;

public class BunnyEars {

static int e=0;
public static int CountEars(int i) {
    if (i == 0) {
        return e;
    } else if (i > 0) {
        if (i % 2 == 0) {
            e = e + 2;
            //i = i - 1;
            CountEars(i-1);
        } else {
            e = e + 3;
            //i = i - 1;
            CountEars(i-1);
        }
    }
    return e;
}

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);

    int b;
    int result;

    System.out.println("Bunny Ears, even has 2 ears, odd has 3 ears");
    System.out.println("Please enter a value: ");
    b = scan.nextInt();

    result = CountEars(b);

    System.out.println("Number of ears are: " + result);
}
}