[week 10][Leetcode][Dynamic Programming] Climbing Stairs
来源:互联网 发布:acfunfix.js 编辑:程序博客网 时间:2024/05/01 21:47
- Question:
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.
- Analysis:
这是一个爬梯子的问题,要到达第n级阶梯则只能是从n-1级阶梯一步跨上来或者是从n-2级阶梯一次跨两步上来,由此很容易可以得到状态转移方程为:
S[n] = S[n-1] + S[n-2]。代码如下所示:
- Code:
class Solution {public: int climbStairs(int n) { int result = 0; int temp[n+1] = {1}; temp[1] = 1; for (int i=2;i<=n;i++) { temp[i] = temp[i-1] + temp[i-2]; } return temp[n]; }};
阅读全文
0 0
- [week 10][Leetcode][Dynamic Programming] Climbing Stairs
- LeetCode /Dynamic Programming/Climbing Stairs
- LeetCode - Fibonacci Sequence/Dynamic Programming - Climbing Stairs
- 70. Climbing Stairs dynamic programming
- Dynamic Programming:70. Climbing Stairs
- [week 7][Leetcode][Dynamic Programming]Triangle
- [week 8][Leetcode][Dynamic Programming] Maximum Subarray
- [week 11][Leetcode][Dynamic Programming] House Robber
- [week 12][Leetcode][Dynamic Programming] Unique Paths
- LeetCode: Climbing Stairs
- LeetCode: Climbing Stairs
- [LeetCode]Climbing Stairs
- LeetCode Climbing Stairs
- [Leetcode] Climbing Stairs
- Leetcode: Climbing stairs
- LeetCode Climbing Stairs
- [LeetCode] Climbing Stairs
- leetcode 107: Climbing Stairs
- 如何生动有趣的入门线性代数
- sessionStorage与localStorage
- java在命令行下编译带有package命名空间的项目
- ANDROID N 分屏设置
- mvs设计模式
- [week 10][Leetcode][Dynamic Programming] Climbing Stairs
- MVC浅谈(转载)
- 关于mysql5.7 [Err] 1055 group by
- Centos7 下如何启动 Tomcat
- Response对象和常用API
- [CQOI2015]网络吞吐量
- Maven管理的jar没有发布到WEB-INF/lib下的解决方案
- LeetCode:Convert Sorted List to Binary Search Tr
- javascript var声明变量提升