70. Climbing Stairs
来源:互联网 发布:中世纪 食物 知乎 编辑:程序博客网 时间:2024/06/06 15:49
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.
改题题意为,上楼梯每步可以走1或者2级阶梯,问走完n级阶梯有多少种不同的方法。
采用动态规划的算法,改题主要需要理解一个原理,走n级阶梯的方法是走n-1级和走n-2级阶梯方法的和。即way[n] = way[n-1]+way[n-2];
那么我们可以知道走1级阶梯有一种方法,走2级阶梯有2种方法;
那么我们就可以利用这个公式算出n级阶梯的方法数量了。
class Solution {public: int climbStairs(int n) { int a[10000]; a[0]=1; a[1]=2; for(int i = 2;i<n;i++) { a[i] = a[i-1]+a[i-2]; } return a[n-1]; }};
阅读全文
0 0
- [LeetCode]70.Climbing Stairs
- 70.Climbing Stairs
- LeetCode --- 70. Climbing Stairs
- [Leetcode] 70. Climbing Stairs
- [leetcode] 70.Climbing Stairs
- 70.Climbing Stairs
- 70. Climbing Stairs
- 70.Climbing Stairs
- 70. Climbing Stairs
- [leetCode]70. Climbing Stairs
- 70. Climbing Stairs
- 70. Climbing Stairs LeetCode
- 70. Climbing Stairs
- 70. Climbing Stairs
- 70. Climbing Stairs
- [LeetCode]70. Climbing Stairs
- 【LeetCode】70. Climbing Stairs
- 70. Climbing Stairs
- javascript开发:javascript机制
- Spring配置文件详解
- solr erro:no servers hosting shard
- DeepLearning tutorial ①
- java nio 基础
- 70. Climbing Stairs
- mysql开启远程访问
- 寻找三角形(待调试)
- Hibernate的一级缓存
- javax.servlet.ServletException: Could not resolve view with name 'xxx' in servlet with name
- Eclipse kotlin插件频繁出错的解决办法
- C++11 模板改进
- 网络嗅探针 Sniffing-Dog
- vee-validate的使用个人小结