Climbing Stairs @python

题目分析：上楼梯问题，每次可上1或2个台阶，求出达到给定的层数有多少种方法。
示例：
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Input: 3
Output: 3

注意：这道题不能用递归，会提示超时，所以需要想一个其他办法。

方法一：

思路：
(1)如果起始跳一阶的话，剩余的n-1阶就有 f(n-1) 种跳法；
(2)如果起始跳二阶的话，剩余的n-2阶就有 f(n-2) 种跳法；
所以f(n) = f(n-1) + f(n-2)，实际结果即为斐波纳契数。
斐波那契数列：n=1–>1；n=2–>2 ；n=3–>n=1的结果+n=2的结果；n=4–>n=2的结果+n=3的结果；n=5–>n=3的结果+n=4的结果… …

该题应采用递归的逆向思维，从底层向上算。观察斐波那契数列得出的规律。

class Solution:def climbStairs(self, n):""":type n: int:rtype: int"""p,q=1,2if n<3:return nelse:for i in range(n-2):p,q=q,p+qreturn q

#Climbing Stairs
class Solution:def climbStairs(self, n):p,q =1,2if n<3:return nelse:for i in range(n-2):p,q = q, p+qreturn q

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