原題地址:https://oj.leetcode.com/problems/climbing-stairs/
題意:
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?
解題思路:
爬樓梯問題。經典的動態規划問題。每次上一個台階或者兩個台階,問一共有多少種方法到樓頂。這個實際上就是斐波那契數列的求解。可以逆向來分析問題,如果有n個台階,那么走完n個台階的方式有f(n)種。而走完n個台階有兩種方法,先走完n-2個台階,然后跨2個台階;先走完n-1個台階,然后跨1個台階。所以f(n) = f(n-1) + f(n-2)。
代碼:
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]
