题解 | #跳台阶#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param number int整型 
# @return int整型
#
class Solution:
    def jumpFloor(self , number: int) -> int:
        # write code here
        if number <=2:
            return number
#         dp = [0] * (number+1)
#         dp[1] = 1
#         dp[2] = 2
#         for i in range(3,number+1):
#             dp[i] = dp[i-1] + dp[i-2]
#         return dp[number]
		# 常数空间复杂度
        a = 1
        b = 2
        c = 0
        for i in range(3,number+1):
            c = a + b
            a,b = b,c # 值的交替迭代
        return c