python各种方法实现斐波那契数列

'''斐波那契数列专题 使用现在学习的各种方法'''

# 数组实现
my_list = [1, 1]
for i in range(2, 10):
    my_list.append(my_list[i-2]+my_list[i-1])
print('数组   ', my_list)

# 交换变量
a = b = 1
for i in range(10-2):
    a, b = b, a+b
print('交换变量 ', b)

# 递归与尾递归
def fbi(n):
    if n < 3:
        return 1
    else:
        return fbi(n-2)+fbi(n-1)


def xfbi(n, i, j):
    if n < 3:
        return j
    else:
        return xfbi(n-1, j, i+j)


print('递归   ', fbi(10))
print('尾递归  ', xfbi(10, 1, 1))


# 迭代器


class MylistIter(object):
    def __init__(self, n):
        self.n = n
        self.i = 1
        self.j = 1

    def __next__(self):
        if self.n > 0:
            v = self.i
            self.i, self.j = self.j, self.i + self.j
            self.n -= 1
            return v
        else:
            raise StopIteration

    def __iter__(self):
        return self


class Mylist(object):

    def __init__(self, n):
        self.n = n

    def __iter__(self):
        return MylistIter(self.n)


for i in Mylist(10):
    pass

print('迭代器1   ', i)



class MylistIter2(object):
    def __init__(self):
        self.i = 1
        self.j = 1

    def __next__(self):
        v = self.i
        self.i, self.j = self.j, self.i + self.j
        return v

    def __iter__(self):
        return self


i = 0
myiter = MylistIter2()
while i < 10:
    a = next(myiter)
    i += 1

print('迭代器2  ', a)

# 生成器
def fun():
    i,j = 1, 1
    for n in range(100):
        yield i
        i, j = j, j+i

f = fun()
for i in range(10):
    b = next(f)

print('生成器  ', b)