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<>python中使用递归实现斐波那契数列
python中使用递归实现斐波那契数列
<>先来了解一下
斐波那契数列（Fibonacci sequence），又称黄金分割数列、因数学家莱昂纳多·斐波那契（Leonardoda
Fibonacci）以兔子繁殖为例子而引入，故又称为“兔子数列”，指的是这样一个数列：0、1、1、2、3、5、8、13、21、34、……在数学上，斐波那契数列以如下被以递推的方法定义：F(0)=0，F(1)=1,
F(n)=F(n - 1)+F(n - 2)（n ≥ 2，n ∈ N）在现代物理、准晶体结构、化学等领域，斐波纳契数列都有直接的应用，为此，美国数学会从
1963 年起出版了以《斐波纳契数列季刊》为名的一份数学杂志，用于专门刊载这方面的研究成果。*
使用递归返回前n项的斐波那契数列：
<>func_1(n-2)+func_1(n-1)此代码为本节代码的主要代码
def func_1(n):
if n == 0:
return 0
elif n == 1 or n == 2:
return 1
else:
return func_1(n-2)+func_1(n-1)
假设n取4
return func_1(2)+func_1(3)
func_1(2)带入函数则返回1，func_1(3)带入函数则返回 func_1(3-2)+func_1(3-1)即为
func_1(1)+func_1(2) ，带入函数为1+1 所以得出结论func_1(4)
=func_1(2)+func_1(3)=func_1(2)+func_1(1)+func_1(2)=3
<>刚才的递推函数只能返回第n各值，要想返回前n项值，得在外围建个函数将得出的值一一添加进去
代码实现：
def func(a):
def func_1(n):
if n == 0:
return 0
elif n == 1 or n == 2:
return 1
else:
return func_1(n-2)+func_1(n-1)
list_1 = []
for i in range(a):
list_1.append(func_1(i))
return list_1
print(func(20))
<>祝大家Python学习顺利！
