题目：

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?

思路：

利用DP的方法，一个台阶的方法次数为1次，两个台阶的方法次数为2个。n个台阶的方法可以理解成上n-2个台阶，然后2步直接上最后一步；或者上n-1个台阶，再单独上一步。 公式是S[n] = S[n-1] + S[n-2] S[1] = 1 S[2] = 2

代码：

class Solution {
public:
    int climbStairs(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        if(n <= 2)
        {
            return n;
        }
        else
        {
            int* step = new int[n];
            
            step[0] = 1;
            step[1] = 2;
            
            for(int i = 2; i < n; i++)
            {
                step[i] = step[i-1] + step[i-2];
            }
            return step[n-1];
        }   
    }
};