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));

    }

};

View Code

2. 和斐波那契亚数列差不多，于是用了for循环来代替，2ms。

 1 class Solution {

 2 public:

 3     int climbStairs(int n) {

 4         if(n <= 2) return n;

 5         

 6         int *result = new int[n];

 7         result[0] = 1;

 8         result[1] = 2;

 9         for(int i = 2; i < n; i++)

10             result[i] = result[i-1] + result[i-2];

11             

12         return result[n-1];

13     }

14 };