python3实现斐波那契数列(4种方法)
人气:0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)
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