通过递归求斐波那契数列

求斐波那契数列的第n项：

def fib1(n):
    if n == 1:
        return 0
    elif n ==2:
        return 1
    else:
        return fib1(n-1) + fib1(n-2)

fib1(3)
# 1

求斐波那契数列的前n项：

def fib2(n):
    if n == 1:
        return [0]
    elif n == 2:
        return [0,1]
    else:
        list1 = fib2(n-1)
        list1.append(list1[-1] + list1[-2])
    return list1

fib2(5)
# [0, 1, 1, 2, 3]

通过迭代器实现求斐波那契数列：

def fib3():
    a = 0
    b = 1
    while 1:
        yield a 
        a,b = b,a + b

number_fib = fib3()        #生成器返回一个迭代器
[next(number_fib) for i in range(10)]

#[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]