题解 | #跳台阶#

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        # write code here
        #典型的动态规划/递归问题，设第i次有dp[i]种解法，则dp[i]=dp[i-1]+dp[i-2]
        #由于空间复杂度为O(1),用三个变量保存这三个状态即可
        dp1 = 1
        dp2 = 2
        dp3 = 3
        if number ==1:
            return dp1
        if number ==2:
            return dp2
        for i in range(2,number):
            dp3 = dp2+dp1
            #状态前移
            dp1 = dp2
            dp2 = dp3
        return dp3