题解 | #跳台阶-斐波那契数列#
跳台阶
https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4
# @param number int整型 # @return int整型 class Solution: def jumpFloor(self , number: int) -> int: # write code here # f(n)=f(n-1)+f(n-2), 初始化f(0)=f(1)=1,后续根据公式计算 a = 1 b = 1 res = 1 for i in range(2, number+1): res = a + b a = b b = res return res