# 70. Climbing Stairs

## 題目原文

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
```

## **解題思路**

{% hint style="info" %}

1. n = 1，output = 1
2. n = 2，output = 2
3. n = 3，output = 3
4. n = 4，output = 5
5. n = 5，output = 8
6. 看似很像是費式數列，但注意的是遞迴版本會超過時間。
   {% endhint %}

## 程式解答

Recursive: **Time Limit Exceeded**

```cpp
class Solution 
{
public:
    int climbStairs(int n) 
    {
        int ans = 0;
        
        if (n <= 2)
            return n;
        
        return climbStairs(n - 1) + climbStairs(n - 2);
    }
};
```

Iterative:  **Accepted**

```cpp
class Solution 
{
public:
    int climbStairs(int n) 
    {
        int *dp = new int[n + 1];
            
        dp[0] = dp[1] = 1;
        
        for (int i = 2; i <= n; i++)
            dp[i] = dp[i-1] + dp[i-2];
        return dp[n];
    }
};
```


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