您可以编写一个fibonacci函数，该函数以线性时间运行并且具有恒定的内存占用，因此不需要列表即可保留它们. 这是一个递归版本(但是，如果n足够大，它将只是 *** )

You can write a fibonacci function that runs in linear time and with constant memory footprint, you don't need a list to keep them. Here's a recursive version (however, if n is big enough, it will just ***)

def fib(a, b, n):
    if n == 1:
        return a
    else: 
        return fib(a+b, a, n-1)


print fib(1, 0, 10) # prints 55

此函数仅调用一次(导致大约N调用参数N)，而您的解决方案则调用两次(大约2 ^ N调用参数N).

This function calls itself only once (resulting in around N calls for a parameter N), in contrast with your solution which calls itself twice (around 2^N calls for a parameter N).

这是一个永远不会堆栈溢出的版本，它使用循环而不是递归:

Here's a version that won't ever *** and uses a loop instead of recursion:

def fib(n):
    a = 1
    b = 0
    while n > 1:
        a, b = a+b, a
        n = n - 1
    return a

print fib(100000)

那已经足够快了:

$ time python fibo.py 
3364476487643178326662161200510754331030214846068006390656476...

real    0m0.869s

但是调用fib直到获得足够大的结果并不完美:意甲联赛的第一个数字被多次计算. 您可以计算下一个斐波那契数，并在同一循环中检查其大小:

But calling fib until you get a result big enough isn't perfect: the first numbers of the serie are calculated multiple times. You can calculate the next fibonacci number and check its size in the same loop:

a = 1
b = 0
n = 1
while len(str(a)) != 1000:
    a, b = a+b, a
    n = n + 1
print "%d has 1000 digits, n = %d" % (a, n)