跳台阶

## 方法一： 递归

class Solution:
    def jumpFloor(self, number):
        # write code here
        if number == 2 or number == 1:
            return number

        return self.jumpFloor(number-1) + self.jumpFloor(number-2)

 ## 方法二：循环实现，避免阶数过大造成时间超限

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        # write code here
        if number == 2 or number == 1:
            return number
        n, m = 1, 2
        for i in range(number-2):
            result = m + n
            n, m = m, result
        return result