70. Climbing Stairs

问题描述

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.

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

题目链接：

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思路分析

给出n阶台阶，一次可以上1阶或者2阶，计算上这n阶台阶有多少种走法。使用动态规划的思想，这就是一个斐波那契数列，将每一步结果放在一个数组中，最后返回结果即可。

代码

class Solution {public:    int climbStairs(int n) {        if (n == 0) return 0;        if (n == 1) return 1;        if (n == 2) return 2;        int sum[n];        sum[0] = 1;sum[1] = 2;        for (int i = 2; i < n; i++){            sum[i] = sum[i-1] + sum[i-2];        }        return sum[n-1];    }};

时间复杂度：O(n)

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反思

一开始想到了用DP，但是没有用，还傻乎乎的计算了一下排列组合，还是太麻烦了，斐波那契最好用。