计算斐波那契数至少为n [英] Calculate Fibonacci numbers up to at least n
问题描述
我正在尝试创建一个允许用户输入数字的函数,结果将是一个包含直到输入为止的斐波那契数字的列表,如果输入不在序列中,则该列表将包含一个以上的数字.例如,输入 4
将返回 [0、1、1、2、3、5]
,但是输入 3
将返回> [0,1,1,2,3]
.我已经使用下面的功能做到了这一点:
I am trying to make a function that allows a user to input a number and the result will be a list containing Fibonacci numbers up to the input and one above if the input is not in the series. For example, input of 4
will return [0, 1, 1, 2, 3, 5]
but input of 3
would return [0, 1, 1, 2, 3]
. I have managed to do this using the function below :
def fibonacci(n):
series = [0]
if (n == 0):
pass
else:
series.append(1)
if (n == 1):
pass
else:
while(series[len(series)-1] < n):
newValue = series[len(series)-1] + series[len(series)-2]
series.append(newValue)
print(series)
但是,我现在想能够递归地执行此操作,有什么想法吗?
However, I would now like to be able to do this recursively, any ideas?
推荐答案
以下是可能的解决方法
def fibo_up_to(n):
if n < 2:
return [1,1]
else:
L = fibo_up_to(n-1)
if L[-1] < n:
L.append(L[-1] + L[-2])
return L
这个想法是要返回所有小于 n
的斐波那契数字的列表,您可以要求小于 n-1
的那些斐波那契数字的列表,然后可能只添加一个元素.如果我们将前两个数字定义为 [1,1]
,则从2开始有效.相反,使用 [0,1]
会造成2的问题,因为单个下一个元素是不够的.
the idea is that to return the list of all fibonacci numbers less than n
you can ask for the list of those less than n-1
and then possibly add just one element. This works from 2 on if we define the first two numbers being [1, 1]
. Using [0, 1]
instead creates a problem for 2 because a single next element is not enough.
这段代码的时间效率不是很高(fibonacci是两次递归,这是一个简单的递归),但是会占用大量堆栈空间.
This code is not inefficient on time (fibonacci is a double recursion, this is a simple recursion), but uses a lot of stack space.
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