题解 | #跳台阶#
跳台阶
https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4
# # 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 == 1 or number == 2: return number # dp[i] 表示跳i级阶梯有多少种方式 dp = [0 for i in range(number+1)] dp[1] = 1 dp[2] = 2 for i in range(3, number+1): dp[i] = dp[i - 1]+ dp[i-2] #rint(dp[number]) return dp[number]