斐波那契数列 迭代与递归效率对比

def myfun2(n):
    n1=1
    n2=1
    while n>2:
        n3=n1+n2
        n1=n2
        n2=n3
        n-=1
    return n3

n = int(input(‘> ‘))
re = myfun2(n)
print(re)                  

def myfun(n):
    if n == 1 or n == 2:
        result = 1
        return result
    else:
        return myfun(n-1) + myfun(n-2)


n=int(input("请输入："))
re = myfun(n)
print(re)