斐波那契数列的通项公式推导:https://blog.csdn.net/zhouchangyu1221/article/details/104974955。
class Solution { public: int Fibonacci(int n) { if (n == 0) return 0; int f1 = 1; int f2 = 1; if (n < 3) return 1; int res; int i = 3; for(; i <= n ; i++) { res = f1 + f2; f1 = f2; f2 = res; } return res; } };