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: 递归会导致超时,用动态规划即可,JAVA实现如下: public int climbStairs(int n) { if(n==1) return 1; e…

题目: 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? Note: Given n will be a positive integer. 题解: 爬到第n层的方法要么是从第n-1层一步上来的,要不就是从n-2层2步…

你正在爬楼梯.需要 n 步你才能到达顶部.每次你可以爬 1 或 2 个台阶.你有多少种不同的方式可以爬到楼顶呢?注意:给定 n 将是一个正整数.示例 1:输入: 2输出: 2说明: 有两种方法可以爬到顶端.1.  1 步 + 1 步2.  2 步示例 2:输入: 3输出: 3说明: 有三种方法可以爬到顶端.1.  1 步 + 1 步 + 1 步2.  1 步 + 2 步3.  2 步 + 1 步详见:https://leetcode.com/problems/climbing-stairs/de…

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…

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? Have you met this question in a real interview? Yes Example Given an example n=3 , 1…

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…

Search a 2D Matrix Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer…

Climbing Stairs https://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? 算法描述 (1)这道题其实是一道fibon…

原题 Climbing Stairs 求斐波那契数列的第N项,开始想用通项公式求解,其实一个O(n)就搞定了. class Solution { public: int climbStairs(int n) { ) ; ; ; ; i<n; ++i) { int t = n2; n2 = n1 + n2; n1 = t; } return n2; } }; 还是觉得没有ACM难度,继续做吧.…

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? 思路:题目也比較简单.类似斐波那契. 代码例如以下: public class Solution { public int climb…