题解 | #跳台阶#

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        # write code here
        # base case
        if number <= 3: 
            return number
        # n = either n-2 +2 or n-1+1
        iCondition1 = self.jumpFloor(number-1) # starting from 1 by +1 and afterward
        iCondition2 = iCondition1 - self.jumpFloor(number-3) # starting from 2 by +2 
        return iCondition1 + iCondition2

重复利用 的值，并减少递归次数。