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java 基础编程练习6
阅读量:713 次
发布时间:2019-03-21

本文共 558 字,大约阅读时间需要 1 分钟。

小乐乐走楼梯的方法数遵循斐波那契数列的规律。当n=1时,只有一种方法;当n=2时,有两种方法。对于更大的n,方法数等于前一阶楼梯的方法数加上第二阶楼梯的方法数,这正是斐波那契数列的定义。通过递归计算,我们可以得到小乐乐的方法数。

具体步骤如下:

  • 当n=1时,返回1。
  • 当n=2时,返回2。
  • 否则,递归调用fun(n-1)和fun(n-2)并相加返回结果。
  • 代码如下:

    public class Main {    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        int n = in.nextInt();        System.out.print(fun(n));    }    private static int fun(int n) {        if (n == 1) {            return 1;        } else if (n == 2) {            return 2;        } else {            return fun(n - 1) + fun(n - 2);        }    }}

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