Climbing Stairs https://oj.leetcode.com/problems/climbing-stairs/ 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? 这题比较简单,可以使用动态规划来求解…
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? 这个爬梯子问题最开始看的时候没搞懂是让干啥的,后来看了别人的分析后,才知道实际上跟斐波那契数列非常相似,假设梯子有n层,那么如何爬到第n层呢,因为每次只能怕1或2步,那…
July 28, 2015 Problem statement: 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? The problem is most popular question in the algorit…
一共有n个台阶,每次跳一个或者两个,有多少种走法,典型的Fibonacii问题 class Solution(object): def climbStairs(self, n): if n<0: return 0 if n<2: return 1 first,second = 1,1 for v in range(2,n+1): res = first+second first,second = return res 还有一种,每次可以跳任意阶,有2^(n-1)种跳法…
题目链接 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? 算法1:分析:dp[i]为爬到第i个台阶需要的步数,那么dp[i] = dp[i-1] + dp[i-2], 很容易看出来这是斐波那契数列的公式 …
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? 原题链接:https://oj.leetcode.com/problems/climbing-stairs/ 题目:你在爬楼梯. 须要 n 步才干到顶部. 每次你爬1…
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? 题意:有n阶楼梯,每次可以走一步或者两步,总共有多少种方法. 思路:动态规划.维护一个一维数组dp[n+1],dp[0]为n=0时的情况,dp[ i ]为到达第i阶台阶…
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? Show Tags 这题其实就是斐波那契数列来的. #include <iostream> using namespace std; class Soluti…
class Solution(object): def climbStairs(self, n): """ :type n: int :rtype: int """ if n <= 2: return n last =1 nexttolast = 1 sums = 0 for i in range(2,n+1): sums = last+nexttolast nexttolast = last last = sums return sums…
