基本的Java递归方法 [英] Basic Java Recursion Method

问题描述

我在java中遇到这个基本的递归问题很麻烦;任何指针都会很棒。

I am having a lot of trouble with this basic recursion problem in java; any pointers would be great.

编写一个静态递归方法来打印出
几何序列的第n项：2 ，6,18,54。

"Write a static recursive method to print out the nth term of the geometric sequence: 2, 6, 18, 54."

从我可以收集到的，在代码中的某处，我应该递归地将某些东西乘以3，但我正在努力弄清楚如何做到这一点。我知道我需要终止声明，但什么时候发生？我需要辅助方法吗？

From what I can gather, somewhere in the code I should be recursively multiplying something by 3, but I'm struggling to figure out how to do this. I know I need a termination statement, but when does that occur? Do I need a helper method?

推荐答案

A 递归函数是一个函数，其实现引用自身。下面是一些有趣的例子：

A Recursive Function is a function whose implementation references itself. Below is some funny example:

public class Inception {
   public void dream() {
      boolean enoughDreaming = false;
      //Some code logic below to check if it's high time to stop dreaming recursively
      ...
      ...

      if(!enoughDreaming) {
           dream(); //Dream inside a Dream
      }
   }
}

以及解决问题的方法：

public class GeometricSequence {
    public static void main(String[] args) {
        //Below method parameters - 5 = n, 1 = count (counter), res = result (Nth number in the GP.
        System.out.println(findNthNumber(5, 1, 2));

    }

    public static int findNthNumber(int n, int count, int res) {
        return ((count == n)) ? res : findNthNumber(n, count+1, res *3);
    }
}

编辑：

上面的类使用int，这只适用于小数字（因为Integer Overflow问题）。 class适用于所有类型/数字：

The above class uses "int", which is good only for small numbers (because of Integer Overflow problem). The below class is better for all types/numbers:

public class GeometricSequence {
    public static void main(String[] args) {
        //Below method parameters - 5 = n, 1 = count (counter), res = result (Nth number in the GP.
        System.out.println(findNthNumber(2000, 1, new BigInteger("2")));

    }

    public static BigInteger findNthNumber(int n, int count, BigInteger res) {
        return ((count == n)) ? res : findNthNumber(n, count+1, res.multiply(new BigInteger("3")));
    }
}

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