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

class Solution {

public:

int climbStairs\(int n\) {

    int one=2,two=1,result=2;

    if \(n==0\){

        return 0;

    }

    if \(n==1\){

        return 1;

    }

    if \(n==2\){

        return 2;

    }

    int count=2,temp;

     while\(count<n\){

         result=two+one;

         temp=one;

         one=result;

         two=temp;

         count++;

     }    

    return result;

}

};

这个就是简单的斐波拉切数列，开始用递归做，这样会超时。然后想用一个数组存结果，但是发现没法调用vector，所以后面就是存了前两个数值即可