solution.md

# 爬楼梯

<p>假设你正在爬楼梯。需要 <em>n</em>&nbsp;阶你才能到达楼顶。</p><p>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？</p><p><strong>注意：</strong>给定 <em>n</em> 是一个正整数。</p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong> 2<strong><br />输出：</strong> 2<strong><br />解释：</strong> 有两种方法可以爬到楼顶。1.  1 阶 + 1 阶2.  2 阶</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong> 3<strong><br />输出：</strong> 3<strong><br />解释：</strong> 有三种方法可以爬到楼顶。1.  1 阶 + 1 阶 + 1 阶2.  1 阶 + 2 阶3.  2 阶 + 1 阶</pre>
<p>以下<span style="color:red">错误</span>的选项是？</p>

## aop

### before

```cpp
#include <bits/stdc++.h>
using namespace std;

```

### after

```cpp
int main()
{
    Solution sol;

    int x = 3;
    int res;
    res = sol.climbStairs(x);
    cout << res;
    return 0;
}
```

## 答案

```cpp
class Solution
{
public:
    int climbStairs(int n)
    {
        double sqrt5 = sqrt(5);
        double fibn = pow((1 + sqrt5) / 2, n) - pow((1 - sqrt5) / 2, n + 1);
        return (int)(fibn / sqrt5);
    }
};
```
## 选项


### A

```cpp
class Solution
{
public:
    int climbStairs(int n)
    {
        vector<int> s;
        s.push_back(1);
        s.push_back(2);
        if (n == 1)
            return 1;
        if (n == 2)
            return 2;
        for (int i = 2; i < n; i++)
        {
            s.push_back(s[i - 1] + s[i - 2]);
        }
        return s[n - 1];
    }
};
```

### B

```cpp
class Solution
{
public:
    int climbStairs(int n)
    {
        vector<int> res(n + 1, 0);
        res[1] = 1;
        res[2] = 2;
        for (int i = 3; i <= n; i++)
            res[i] = res[i - 1] + res[i - 2];
        return res[n];
    }
};
```

### C

```cpp
class Solution
{
public:
    int climbStairs(int n)
    {
        if (n == 1)
        {
            return 1;
        }
        int first = 1;
        int second = 2;
        for (int i = 3; i <= n; i++)
        {
            int third = first + second;
            first = second;
            second = third;
        }
        return second;
    }
};
```