LN : leetcode 70 Climbing Stairs
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.
DP Accepted
这是一道特别经典又简单的动态规划题目,先考虑简单的情况,上第一层的方式只有1种,而上第二层的方式有两种。再考虑复杂的情况,上第n层的前一层要么是n-1层,要么是n-2层。其实,该问题很类似斐波那契数列,还是很容易解答的。
class Solution {
public:
int climbStairs(int n) {
vector<int> stairs(n+1, 0);
stairs[1] = 1;
stairs[2] = 2;
for (int i = 3; i < n+1; i++) stairs[i] = stairs[i-1]+stairs[i-2];
return stairs[n];
}
};
LN : leetcode 70 Climbing Stairs的更多相关文章
- 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 ...
- [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
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 ...
- LeetCode 70 Climbing Stairs(爬楼梯)(动态规划)(*)
翻译 你正在爬一个楼梯. 它须要n步才干究竟顶部. 每次你能够爬1步或者2两步. 那么你有多少种不同的方法爬到顶部呢? 原文 You are climbing a stair case. It tak ...
- Java [Leetcode 70]Climbing Stairs
题目描述: You are climbing a stair case. It takes n steps to reach to the top. Each time you can either ...
- [leetcode]70. Climbing Stairs爬楼梯
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 (C++)
题目: You are climbing a stair case. It takes n steps to reach to the top. Each time you can either cl ...
随机推荐
- hdu1576 mod 运算的逆元
Problem Description 要求(A/B)%9973,但因为A非常大,我们仅仅给出n(n=A%9973)(我们给定的A必能被B整除.且gcd(B,9973) = 1). Input 数 ...
- STM32通过调用库函数进行编程
1.调用库函数编程和直接配置寄存器编程的差别: 2.CMSIS标准: 3.STM32库函数的组织: 4.程序例举: 调用库函数实现通过USART发送数据(26个大写的英文字母) 首先:在主函数部分先要 ...
- RabbitMQ常用命令、管理界面
1.运行CMD,cd切换到RabbitMQ安装目录sbin下E:\Program Files\RabbitMQ Server\rabbitmq_server-3.7.2\sbin 执行 rabbitm ...
- HDU 5274 Dylans loves tree 树链剖分+线段树
Dylans loves tree Problem Description Dylans is given a tree with N nodes. All nodes have a value A[ ...
- struts2的一些小问题
1.action和ValueStack的关系2.ValueStack的类set()方法和setValue()方法的区别3.ValueStack的类push()方法的作用4.从ValueStack对象中 ...
- vc字符串转换处理:(绝对精华,收集所有的例子)
vc字符串转换处理:(绝对精华,收集所有的例子) 1.头文件中要定义宏; #define UNICODE #define _UNICODE //////////// ...
- HDU3488 Tour —— 二分图最大权匹配 KM算法
题目链接:https://vjudge.net/problem/HDU-3488 Tour Time Limit: 3000/1000 MS (Java/Others) Memory Limit ...
- linux环境下oracle静默安装
一.安装环境 1.linux版本:redhat6.3_x86_64 2.oracle版本:Oracle Database 11g Enterprise Edition Release 11.2.0.3 ...
- 基于 IOCP 的通用异步 Windows Socket TCP 高性能服务端组件的设计与实现
设计概述 服务端通信组件的设计是一项非常严谨的工作,其中性能.伸缩性和稳定性是必须考虑的硬性质量指标,若要把组件设计为通用组件提供给多种已知或未知的上层应用使用,则设计的难度更会大大增加,通用性.可用 ...
- bzoj4811 [Ynoi2017]由乃的OJ 树链剖分+位运算
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=4811 因为位运算的结果有可合并性,所以可以树链剖分,线段树维护: 细节很多,特别要注意从左往 ...