[leetcode]Climbing Stairs @ Python

原题地址：https://oj.leetcode.com/problems/climbing-stairs/

题意：

You are climbing a stair case. It takes n steps to reach
to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can
you climb to the top?

解题思路：

爬楼梯问题。经典的动态规划问题。每次上一个台阶或者两个台阶，问一共有多少种方法到楼顶。这个实际上就是斐波那契数列的求解。可以逆向来分析问题，如果有n个台阶，那么走完n个台阶的方式有f(n)种。而走完n个台阶有两种方法，先走完n-2个台阶，然后跨2个台阶；先走完n-1个台阶，然后跨1个台阶。所以f(n)
= f(n-1) + f(n-2)。

代码：

class Solution:

    # @param n, an integer

    # @return an integer

    def climbStairs(self, n):

        dp = [1 for i in range(n+1)]

        for i in range(2, n+1):

            dp[i] = dp[i-1] + dp[i-2]

        return dp[n]

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