算法设计Week8 LeetCode Algorithms Problem #70 Climbing Stairs
2017-04-16 22:00
387 查看
题目描述:
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.
题目分析:
使用动态规划算法,设climb[i]为爬上第i级台阶不同方法的总数,那么有两种方式可以爬上第i级台阶:1. 爬上第i-1级台阶后再向上爬一级台阶;
2. 爬上第i-2级台阶后再向上爬两级台阶。
这样,就可以得到递推式:
climb[i] = climb[i - 1] + climb[i - 2].
此外,climb[1] = 1, climb[2] = 2。
具体代码如下(注意代码中climb数组下标从0开始),算法的时间复杂度为O(n),空间复杂度为O(n)。
class Solution { public: int climbStairs(int n) { int climb = {0}; climb[0] = 1; for(int i = 1; i < n; i++){ if(i == 1) climb[i] = 2; else climb[i] = climb[i - 1] + climb[i - 2]; } return climb[n - 1]; } };
相关文章推荐
- 算法设计Week8 LeetCode Algorithms Problem #198 House Robber
- Leetcode 70 Climbing Stairs
- LeetCode70 Climbing Stairs
- leetcode 70 Climbing Stairs
- LeetCode-70-Climbing Stairs
- leetcode——70——Climbing Stairs
- LeetCode(70) Climbing Stairs
- LeetCode-70 climbing stairs(方法大总结)
- [Leetcode 29] 70 Climbing Stairs
- LeetCode 70 Climbing Stairs(爬楼梯)(动态规划)(*)
- LeetCode 70 — Climbing Stairs(C++ Java Python)
- LeetCode-70-Climbing Stairs(爬楼梯)
- LeetCode 70 Climbing Stairs
- LeetCode(70)Climbing Stairs
- [leetcode 70]Climbing Stairs
- LeetCode 70 Climbing Stairs--Python实现
- LeetCode-70-Climbing Stairs-E
- Climbing Stairs | leetcode 70 【Java解题报告】
- leetcode[70]:Climbing Stairs
- LeetCode70 Climbing Stairs 解题报告