Question

即使有记忆，我也无法计算斐波那契（1000）。这是我的代码（与关于memoization的教科书示例相同）。

Flask-SQLAlchemy==2.2
Flask==0.12.2
SQLAlchemy==1.1.13

我得到的错误：

inputnumber = 10000 # the error starts if n > 998
existingvalues = {}

def fib_series(n):
    if(n <= 1):
        return n
    else:
        if(n in existingvalues):
            value = existingvalues[n]
        else:
            value = fib_series(n-1) + fib_series(n-2)
            existingvalues[n] = value
        return value

我想要这样做的原因是比较递归和迭代之间的性能。以下代码进行迭代并且可以找到n>的fibonacci。 1000在几乎相似的时间。

Traceback (most recent call last):
  File "C:/PycharmProjects/learnpy/11 - recursion.py", line 24, in <module>
    print(fib_series(inputnumber))
  File "C:/PycharmProjects/learnpy/11 - recursion.py", line 18, in fib_series
    value = fib_series(n-1) + fib_series(n-2)
  File "C:/PycharmProjects/learnpy/11 - recursion.py", line 18, in fib_series
    value = fib_series(n-1) + fib_series(n-2)
  File "C:/PycharmProjects/learnpy/11 - recursion.py", line 18, in fib_series
    value = fib_series(n-1) + fib_series(n-2)
  [Previous line repeated 994 more times]
  File "C:/PycharmProjects/learnpy/11 - recursion.py", line 12, in fib_series
    if(n <= 1):
RecursionError: maximum recursion depth exceeded in comparison

在比较这两种加法技术（递归与迭代）时，是否有与O（复杂性）相关的内容？

用于n的Python中的Fibonacci级数计算> 1000

0 个答案: