LeetCode - Fibonacci Sequence/Dynamic Programming - 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?

f (n) = f (n-1) + f (n-2)，其中f(0)=1，f(-1)=0

class Solution {
public:
    int climbStairs(int n) {
        if(n==0)
            return 0;
        int prev=0, cur=1;
        for(int i=1;i<=n;i++)
        {
            int temp=cur;
            cur+=prev;
            prev=temp;
        }
        return cur;
    }
};

Fibonacci Sequence计算公式

class Solution {
public:
    int climbStairs(int n) {
        double s=sqrt(5);
        return floor((pow((1+s)/2,n+1)-pow((1-s)/2,n+1))/s);
    }
};