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…

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? 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…

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…

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step w…

目录 题目链接 注意点 解法 小结 题目链接 Min Cost Climbing Stairs - LeetCode 注意点 注意边界条件 解法 解法一:这道题也是一道dp题.dp[i]表示爬到第i层的最小cost,想要到达第i层只有两种可能性,一个是从第i-2层上直接跳上来,一个是从第i-1层上跳上来.所以可以得到dp[i] = min(dp[i- 2] + cost[i - 2], dp[i - 1] + cost[i - 1]).时间复杂度O(n). class Solution { pu…