10. 斐波那契数列
斐波那契数列
http://www.nowcoder.com/questionTerminal/c6c7742f5ba7442aada113136ddea0c3
- 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) - 利用矩阵
https://blog.csdn.net/lamusique/article/details/89161831 - 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]
感觉像是高情商回复
