斐波那契数的迭代算法 [英] An iterative algorithm for Fibonacci numbers

问题描述

我对斐波那契数的迭代算法感兴趣，因此我在Wiki上找到了公式...它看起来很直接，所以我在Python中尝试了它...编译没有问题，公式看起来也不错。 ..不确定为什么会给出错误的输出...我执行不正确吗？

I am interested in an iterative algorithm for Fibonacci numbers, so I found the formula on wiki...it looks straight forward so I tried it in Python...it doesn't have a problem compiling and formula looks right...not sure why its giving the wrong output...did I not implement it right ?

def fib (n): 
    if( n == 0):
        return 0
    else:
        x = 0
        y = 1
        for i in range(1,n):
            z = (x + y)
            x = y
            y = z
            return y

for i in range(10):
    print (fib(i))

输出

0

无

1

1

1

1

1

1

0
None
1
1
1
1
1
1

推荐答案

问题是您的返回y 在您的功能循环。因此，在第一次迭代之后，它将已经停止并返回第一个值：1.除非 n 为0时，否则函数将返回 0 本身，并且在 n 为1的情况下，for循环甚至不会迭代一次，并且没有 return 正在执行（因此 None 返回值）。

The problem is that your return y is within the loop of your function. So after the first iteration, it will already stop and return the first value: 1. Except when n is 0, in which case the function is made to return 0 itself, and in case n is 1, when the for loop will not iterate even once, and no return is being execute (hence the None return value).

要解决此问题，只需将返回y 移出循环即可。

To fix this, just move the return y outside of the loop.

以KebertX的示例为例，这是我亲自在Python中制作的解决方案。当然，如果要处理许多斐波那契值，您甚至可能希望将这两种解决方案结合起来并为数字创建缓存。

Following KebertX’s example, here is a solution I would personally make in Python. Of course, if you were to process many Fibonacci values, you might even want to combine those two solutions and create a cache for the numbers.

def f(n):
    a, b = 0, 1
    for i in range(0, n):
        a, b = b, a + b
    return a

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