题解 | #跳台阶#

迭代：空间占用过多

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        if number  <= 2:
            return number
        else:
            dp = [0]*(number+1)
            dp[0] = 0
            dp[1] = 1
            dp[2] = 2

            for i in range(3,number+1):
                dp[i] = dp[i-1] + dp[i-2]

            return dp[number]

减少空间：只用两个变量来存

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        if number  <= 2:
            return number
        else:
            pre1, pre2 = 1, 2

            for i in range(3,number+1):
                cur = pre1 + pre2
                pre1 = pre2
                pre2 = cur

            return pre2

递归：n很大时超时

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        if number <= 2:
            return number
        else:
            return self.jumpFloor(number-1) + self.jumpFloor(number-2)