41-Climbing Stairs-leetcode

1. Climbing Stairs My Submissions QuestionEditorial Solution
   Total Accepted: 106498 Total Submissions: 290003 Difficulty: Easy
   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：so easy
重要的是思路清晰，对于n阶楼梯，可能由n-1阶迈上来的，也可能由n-2阶楼梯迈上来的，

记f(n)为迈上n阶楼梯所需的步数

f(n)=f(n−1)+f(n−2)

f(2)=2,f(1)=1

就是一斐波纳挈数列，通项和为

an=15√((1+5√2)n−(1−5√2)n)

这里的 f(n)=an+1
这里程序有三种实现方式，递归效率最差，自底向上循环其次，最好的是常数级，用公式

class Solution {
public:
    int climbStairs(int n) {
        vector<int> vec(n+1,0);
        vec[1]=1;vec[2]=2;
        if(n<=2)return vec[n];
        for(int i=3;i<=n;++i)
            vec[i]=vec[i-1]+vec[i-2];
        return vec[n];
    }
};