[LC] 70. 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?
Note: Given n will be a positive integer.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
class Solution {
public int climbStairs(int n) {
if (n <= 2) {
return n;
}
int preOne = 2;
int preTwo = 1;
int sum = 0;
for (int i = 3; i <= n; i++) {
sum = preOne + preTwo;
preTwo = preOne;
preOne = sum;
}
return sum;
}
}
[LC] 70. Climbing Stairs的更多相关文章
- LN : leetcode 70 Climbing Stairs
lc 70 Climbing Stairs 70 Climbing Stairs You are climbing a stair case. It takes n steps to reach to ...
- 42. leetcode 70. Climbing Stairs
70. Climbing Stairs You are climbing a stair case. It takes n steps to reach to the top. Each time y ...
- Leetcode#70. Climbing Stairs(爬楼梯)
题目描述 假设你正在爬楼梯.需要 n 阶你才能到达楼顶. 每次你可以爬 1 或 2 个台阶.你有多少种不同的方法可以爬到楼顶呢? 注意:给定 n 是一个正整数. 示例 1: 输入: 2 输出: 2 解 ...
- 377. Combination Sum IV 70. Climbing Stairs
back function (return number) remember the structure class Solution { int res = 0; //List<List< ...
- Leetcode之70. Climbing Stairs Easy
Leetcode 70 Climbing Stairs Easy https://leetcode.com/problems/climbing-stairs/ You are climbing a s ...
- 刷题70. Climbing Stairs
一.题目说明 题目70. Climbing Stairs,爬台阶(楼梯),一次可以爬1.2个台阶,n层的台阶有几种爬法.难度是Easy! 二.我的解答 类似的题目做过,问题就变得非常简单.首先用递归方 ...
- LeetCode练题——70. Climbing Stairs
1.题目 70. Climbing Stairs——Easy You are climbing a stair case. It takes n steps to reach to the top. ...
- [LeetCode] 70. Climbing Stairs 爬楼梯问题
You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb ...
- leetCode 70.Climbing Stairs (爬楼梯) 解题思路和方法
Climbing Stairs You are climbing a stair case. It takes n steps to reach to the top. Each time you ...
随机推荐
- ZJNU 2356 - 六学家
“选出来三个六学家,他们的编号是i,j,k,满足i<j<k,且a[k]=a[j]-a[i]” 所以输入第i个数a[i]时,直接让答案加上前i-1个数中能构成差值为a[i]的数量即可 然后让 ...
- ubuntu下安装ant
背景介绍 最近终于正式开始填补一下自己在web方面的知识漏洞. 而ant则是必不可少的东西了,要问ant的作用是什么,简单的说,这个软件可以用最简单的方法将你的web应用程序部署到服务器上,是不是很强 ...
- vue打包空白及字体路径错误问题
vue项目打包后空白 在config/index.js文件中 assetsPublicPath: '/', 改为 assetsPublicPath: './', build: { // Templat ...
- 小结spring给项目开发的好处
1.spring 抽象了许多开发中遇到的共性问题:支持pojo和javaBean开发使应用面向接口开发.如各种Template 2.Ioc 容器使得对象间的耦合关系文本化.外部化,即通过xml的配置就 ...
- Ribbon使用及其客户端负载均衡实现原理分析
1.ribbon负载均衡测试 (1)consumer工程添加依赖 <dependency> <groupId>org.springframework.cloud</gro ...
- 从定时器的选型,到透过源码看XXL-Job(上)
此内容来自一位好朋友的分享,也是当初建议我写博客提升的朋友.内容只做转载,未做修改. 定时任务选型 背景 目前项目定时任务采用Spring Task实现,随着项目需求的迭代,新增的定时任务也越来越多. ...
- mac下使用opencv编译安装新模块contrib
opencv-4.0.1 opencv_contrib-4.0.1 提供ippicv下载链接: https://pan.baidu.com/s/1OIJRUqPqAtpMetku8qX36w cont ...
- 在项目中ES6语法转ES5语法 babel转码器
es6 babel 安装以及使用 1,安装好node(需要使用npm包管理工具) 2,在本地项目路径下npm init,格式化成功后会在项目下生成一个配置文件package.json 3,本地安装 ...
- Java基础二(2020.1.14)
学习内容: 1.Java运算符:赋值运算符,算术运算符,关系运算符,逻辑运算符,条件运算符 2.java流程控制:顺序,选择 1.java输入 Scanner s = new Scanner(Syst ...
- 推荐:MongoDB学习资料
http://www.mongodb.org/display/DOCS/Production+Deployments Official MongoDBProject Website Getting S ...