两个斐波拉切数列运算的时间复杂度 顶

import java.util.Scanner;

/**
 * Created by Administrator on 2018-02-15.
 */
public class Fibonacci {
    public static long FibonacciQty(int n) {
        long t1,t2;
        if(n == 1 || n == 2) {
            return 1;
        }else {
            t1 = FibonacciQty(n-1);
            t2 = FibonacciQty(n-2);
            return t1+t2;
        }
    }
    public static long FibonacciQty2(int n) {
        long t1 = 1;
        long t2 = 1;
        if(n == 1 || n == 2) {
            return 1;
        }else {
            for (int i = 2;i < n - 1;i++) {
                t2 = t1 + t2;
                t1 = t2 - t1;
            }
            return t1+t2;
        }
    }
    public static void main(String[] args) {
        System.out.println("递推算法求解兔子产仔问题!");
        System.out.print("请先输入时间:");
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        long num = Fibonacci.FibonacciQty2(n);
        System.out.println("经过"+n+"月的时间，共能繁殖成"+num+"对兔子!");
    }
}