1. 递归，为了提高效率，我们需要利用一个辅助的结果列表result，计算过的n和Fibonacci(n)对储存在里面，避免重复计算
    # -*- coding:utf-8 -*-
class Solution:
 def Fibonacci(self, n):
     # write code here
     result = {0:0, 1:1}
     def helper(n):
         if n in result:
             return result[n]
         res = helper(n-1) + helper(n-2)
         result[n] = res
         return res
     return helper(n)
  2. 利用矩阵
    https://blog.csdn.net/lamusique/article/details/89161831
  3. 更简单实用的是用循环来做
    class Solution:
 def Fibonacci(self, n):
     # write code here
     res = [0,1,1,2]
     while len(res)-1  < n:
         res.append(res[-1]+res[-2])
     return res[n]