斐波那契数列Python实现

def fbi(n):
    s1 = 1
    s2 = 1
    if(n<=0):
        return -1
    else:
        if( n==1 | n==2):
            return 1
        else:
            for num in range(2, n):  
                s2 = s1+s2
                s1 = s2-s1
            return s2


print(fbi(9))

时间复杂度为o(n),空间复杂度为o(1),用的是迭代法想加

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转载自blog.csdn.net/huakaiba/article/details/82109779