Climbing Stairs

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?

 

分析：考虑走第n步时的情况，可以从第n-1个台阶走一步，也可以从第n-2个台阶走两步。即f(n) = f(n-1) + f(n-2)，同时f(1) = 1, f(2) = 2.

 

1. 使用递归，结果超时了。

class Solution {
public:int climbStairs(int n) {if(n == 1) return 1;if(n == 2) return 2;else return(climbStairs(n-1) + climbStairs(n-2));}
};

2. 和斐波那契亚数列差不多，于是用了for循环来代替，2ms。

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