python3实现斐波那契数列--4种方法

• 基础版(list方法)

# 比较占内存
w = int(input("输入一个数字还你一个斐波那契数列："))
list_res = []
def list_n(n):
    if n>=3:
        res=list_n(n-1)+list_n(n-2)
    else:
        res=1
    return res

print("开始")

for i in range(0,w):
    list_res.append(list_n(i+1))
print(list_res)

• 升级版

# 比较占内存
num =int(input("输入一个数字还你一个斐波那契数列v2.0："))
list_nums=[1,1]
def calculate(num,list_nums):
    i = 0
    if num>2:
        while i < num:
            list_nums.insert(i+2,list_nums[i]+list_nums[i+1])
            i+=1
        else:
            print("数列已生成")
            print(list_nums)
        return list_nums[num-1]
    else:
        return list_nums[0]

res = calculate(num,list_nums)
print("="*50)
print("第%s个:%s"%(num,res))

• 最实用版(解包的方式)

#省内存
def fbnq(n):
    a,b=1,1
    if n==1 or n ==2:
        return 1
    else:
        i=3
        while i<=n:
            a,b=b,a+b
            i+=1
        return b

print(fbnq(int(input("输入一个数："))))

• 迭代器版

"""实现斐波那契数列"""


class feibo(object):
    def __init__(self, length):
        self.num1 = 0
        self.num2 = 1
        self.num = self.num1
        self.length = length
        self.index = 0

    def __iter__(self):
        return self

    def __next__(self):
        self.num = self.num1
        while True:
            if self.index == self.length:
                raise StopIteration
            self.num1, self.num2 = self.num2, self.num1+self.num2
            self.index += 1
            return self.num


myfbnq = feibo(10)
# print(list(myfbnq)) # 指针位置已到最后一位
for i in myfbnq:
    print(i)