变态跳台阶

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloorII(self, number):
        #number=1 1
        #numeber=2 2
        #number=3 4
        #number=4 8
        #number=n 2^(n-1)
        #f（n）=f(n-1)+f(n-2)+...f(0)
        #f(n-1)=f(n-2)+...+f(0)
        #f(n)=2f(n-1) n>1
        #f(1)=1 n=1
        if number ==1:
            return 1
        ret = 1
        a=1
        for i in range(2,number+1):
            ret=2*a
            a=ret
        return ret
        #return pow(2,number-1)找规律的方法
        # write code here