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? 思路:这道题是斐波那契数列的延伸.首先用最简单的递归的方法 class Solution { public: int climbStairs(int n) { ) re…

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…

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…

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…

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? 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…

最近在准备找工作的算法题,刷刷LeetCode,以下是我的解题报告索引,每一题几乎都有详细的说明,供各位码农参考.根据我自己做的进度持续更新中......                              本文地址 LeetCode:Reverse Words in a String LeetCode:Evaluate Reverse Polish Notation LeetCode:Max Points on a Line LeetCode:Sort List LeetCode:Ins…