跳台阶

#include <iostream>

class Solution {
public:
    int jumpFloor(int number) {
        if (number == 1 || number == 2)
        {
            return number;
        }
        
        //典型的斐波那契数列问题，但完全用递归效率极低，所以这里替换为循环
        /*else
        {
            return (jumpFloor(number - 1) + jumpFloor(number - 2));
        }*/
        
        long int Fn = 0, Fn_1 = 2, Fn_2 = 1;
        for (; number >= 3; number--)
        {
            
            Fn = Fn_1 + Fn_2;
            Fn_1 = Fn;
            Fn_2 = Fn_1;
        }
        return Fn;
    }
};

void main(){
    Solution* a = new Solution;
    std::cout<<a->jumpFloor(30);
}