题解 | #跳台阶#

# 动态规划
class Solution:
    def jumpFloor(self , number: int) -> int:
        # write code here
        dp = [0 for _ in range(number+1)] # dp[i] 表示跳上第i个台阶的总跳法
        dp[0], dp[1] = 1, 1 # 初始化
        for i in range(2, number+1):
            dp[i] = dp[i-1] + dp[i-2]
        return dp[number]