该程序本身运行良好.输入的每个值都将返回正确的对应斐波那契数.我还需要定时执行递归函数和使用循环的函数.当我运行程序时,它只是返回用户实际输入其输入所花费的时间,而不是运行函数所花费的时间.我已经尝试过移动计时器,但我一直得到相同的结果.我想返回递归计算斐波那契数的实际时间,以及使用循环计算斐波那契数的时间.

import java.util.Scanner;

public class FibNumbers {
            public static void main(String[] args) {
            int f;
            int l;
            long num1;
            long num2;
            int choice;
            long t1 = System.currentTimeMillis();
            Scanner sc = new Scanner(System.in);
            Scanner keyboard = new Scanner(System.in);

            System.out.println("For recursion press 1 %n for loop press 2" );
            choice = keyboard.nextInt();
            if(choice == 1) {
                System.out.println("Please enter a value for n");
                f = Integer.parseInt(sc.nextLine());
                System.out.println("The Fibonacci number is " + fibonacciR(f) + " using recursion");
            }
            else {
                System.out.println("Please enter a value for n");
                l = Integer.parseInt(sc.nextLine());
                System.out.println("The Fibonacci number is " + fibonacciL(l) + " using a for loop");
            }
            long t2 = System.currentTimeMillis();
            System.out.println("The elapsed time is " + (t2 - t1) / 1000 + " seconds.");
        }

        public static long fibonacciR(long f) {
            if(f == 0) {
                return 0;
            }
            else if(f == 1) {
                return 1;
            }
            else {
                return fibonacciR(f-1) + fibonacciR(f-2);
            }
        }

        public static long fibonacciL(long l) {
            long num1 = 0;
            long num2 = 1;
            long total = l;
            long sumOfTwo;
            for(int i = 0; i < total; i++) {
                sumOfTwo = num1 + num2;
                num1 = num2;
                num2 = sumOfTwo;
            }
            return num1;
        }
    }

解决方法:

使用System.nanoTime();纳秒.

如果(t2-t1)小于1000,则除以1000将得到0,因为您使用的时间很长,可以将其强制转换为float / double,然后进行除法.

可能是这样的：

import java.util.Scanner;

public class FibNumbers {
        public static void main(String[] args) {
        int f;
        int l;
        long num1;
        long num2;
        int choice;
        long t1 = 0, t2 = 0;
        Scanner sc = new Scanner(System.in);
        Scanner keyboard = new Scanner(System.in);

        System.out.println("For recursion press 1 %n for loop press 2" );
        choice = keyboard.nextInt();
        if(choice == 1) {
            System.out.println("Please enter a value for n");
            f = Integer.parseInt(sc.nextLine());
            t1 = System.nanoTime();
            System.out.println("The Fibonacci number is " + fibonacciR(f) + " using recursion");
            t2 = System.nanoTime();
        }
        else {
            System.out.println("Please enter a value for n");
            l = Integer.parseInt(sc.nextLine());
            t1 = System.currentTimeMillis();
            System.out.println("The Fibonacci number is " + fibonacciL(l) + " using a for loop");
            t2 = System.nanoTime();
        }
        System.out.println("The elapsed time is " + (t2 - t1) + " nano seconds.");
    }

    public static long fibonacciR(long f) {
        if(f == 0) {
            return 0;
        }
        else if(f == 1) {
            return 1;
        }
        else {
            return fibonacciR(f-1) + fibonacciR(f-2);
        }
    }

    public static long fibonacciL(long l) {
        long num1 = 0;
        long num2 = 1;
        long total = l;
        long sumOfTwo;
        for(int i = 0; i < total; i++) {
            sumOfTwo = num1 + num2;
            num1 = num2;
            num2 = sumOfTwo;
        }
        return num1;
    }
}