python 斐波那契数列实现方法

1. 
   # while

   def Fibonacci1(num):
       a,b=0,1
       while a<num:
           print(a)
           a,b=b,a+b
   # print(Fibonacci1(20))

    

2. # yield
   

   def Fibonacci2(num):
       a,y,b=0,1,1
       while a<num:
           yield y
           y,b=b,y+b
           a=a+1
   res=Fibonacci2(10)
   fbn=[i for i in res]
   # print(fbn)
   # [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]

    

3. for
   

   def Fibonacci3(n):
       a,b=0,1
       for i in range(n):
           a,b=b,a+b
           print(a)
   # print(Fibonacci3(10))

    

# 上述的递归在时间复杂度上面，效率较慢，不推荐

# 矩阵法 时间复杂度O(log n)

import numpy

def fib_matr(n):
    return (numpy.matrix([[1, 1], [1, 0]]) ** (n - 1) * numpy.matrix([[1], [0]]))[0, 0]

for i in range(20):
    print(int(fib_matr(i)), end=" ")
# 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181

 

来源：CSDN

作者：小弧y

链接：https://blog.csdn.net/weixin_38128649/article/details/104008152