对于这道题，F(n)=1+F(n-1)+F(n-2)+ ... +F(1)
其中F(1)=1,F(2)=2
本题可用循环实现： 
# -*- coding:utf-8 -*-
class Solution:
    def jumpFloorII(self, n):
        if n==0:
            return 0
        if n==1:
            return 1
        if n==2:
            return 2
        elif n>2:
            result=(n+1)*[0]
            result[1]=1
            result[2]=2
            sum=result[0]+result[1]+result[2]
            for i in range(3,n+1):
                result[i]=sum+1
                sum+=result[i]
            return result[n]
        # write code here