[牛客网,剑指offer,python] 跳台阶
2017-07-02 20:03
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跳台阶
题目描述
一只青蛙一次可以跳上1级台阶,也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法。解题思路
当前台阶的跳法种数 = 当前台阶后退一个台阶的跳法总数 + 当前台阶后退两个台阶的跳法总数即:f (n) = f (n-1) + f (n-2)
问题则变得和斐波那契数列很相似了,但初值有一点区别。
代码
class Solution: def jumpFloor(self, number): # write code here if number < 3: return number a = [0,1,2] for i in range(3, number+1): a.append(a[i-1] + a[i-2]) return a[number]
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