# 70.Climbing Stairs

**70.Climbing Stairs**

难度:Easy

> 假设你正在爬楼梯。需要 n 阶你才能到达楼顶。

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？

注意：给定 n 是一个正整数。

```
示例 1：

输入： 2
输出： 2
解释： 有两种方法可以爬到楼顶。
1.  1 阶 + 1 阶
2.  2 阶
示例 2：

输入： 3
输出： 3
解释： 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
```

类似斐波那契数列解法。

```
class Solution {
    vector<int>num={1,2,3};
public:
    int climbStairs(int n) {
        if(n<=num.size()) return num[n-1];
        int res= climbStairs(n-1)+ climbStairs(n-2);
        num.push_back(res);
        return res;
    }
};
```

> 执行用时 :0 ms, 在所有 C++ 提交中击败了100.00%的用户\
> 内存消耗 :8.7 MB, 在所有 C++ 提交中击败了9.73%的用户


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://dfine.gitbook.io/leetcode/70.climbing_stairs.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.