题解 | #跳台阶#

到第n个台阶有f(n)种方法，所有可能情况：①从n-1跳一个，②从n-2跳两个，即f(n) = f(n-1) + f(n-2)也就是斐波那契数列。采用递归+缓存，也可以说是动态规划的方案

from functools import lru_cache
class Solution:
    @lru_cache
    def jumpFloor(self , number: int) -> int:
        # write code here
        if number <= 2:
            return number
        return self.jumpFloor(number-1) + self.jumpFloor(number-2)