[LeetCode] Climbing Sairs

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：这道题是斐波那契数列的延伸。首先用最简单的递归的方法

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

不出意料的超时了。替代递归的方法是用动态规划。

class Solution {
public:
    int climbStairs(int n)
    {
        vector<int> res(n+);
        res[] = ;
        res[] = ;
        for (int i = ; i <= n; i++)
        {
            res[i] = res[i-] + res[i-];
        }
        return res[n];
    }  

};

时间复杂度降低了，接下来降低空间复杂度。用变量代替数组

class Solution {
public:
    int climbStairs(int n)
    {
        if (n <= ) return n;
        int f1 = ;
        int f2 = ;
        int f3 = ;
        for (int i = ; i <= n; ++i) {
            f3 = f2 + f1;
            f1 = f2;
            f2 = f3;
        }

        return f3;
    }  

};

最终的时间复杂度O(n),空间复杂度O(1)