Question

我试图将它打印出代码的递归部分，就像它打印出迭代一样。迭代中的斐波那契序列

9时斐波那契序列的结束是34 时间：1。像这样，但每次我尝试循环打印我得到一些奇怪的答案，不确定要添加到方法。

    import java.util.Scanner;

    public class fibonacciRecursive {

    public static void main(String[] args) {
        /*
         * Dustin Willingham 
         * Class: CSCI 1302 Mon-Wed Cartersville 
         *Lets user enter the last number of a fibonacci sequence
         *then it takes the number and runs the fibonacci sequence in recursive and iteration method
         *It outputs the answer for both and the time
         */

        Scanner in = new Scanner(System.in);
        System.out.println("Enter the number you want the Fibonacci sequence to go to:");
        int fibNumb = in.nextInt();

        //Print and timing for the Fibonacci sequence with the Recursive method
        System.out.println("Fibonacci sequence in recursion");
        //start of the timer
        long start = System.currentTimeMillis();
        System.out.println("The ending of the fibonacci sequence at " + fibNumb + " is " + fibonacciRec(fibNumb));
        System.out.print("Time : ");
        //print out the timer
        System.out.println(System.currentTimeMillis() - start);





        //Print and timing for the Fibonacci sequence with the Iteration method
        System.out.println("\n" + "Fibonacci sequence in iteration");
        start = System.currentTimeMillis();
        System.out.println("The ending of the fibonacci sequence at " + fibNumb + " is " + fiboacciIte(fibNumb));
        System.out.print("Time : ");
        System.out.print(System.currentTimeMillis() - start);


        in.close();
    }

    // Recursive method for the Fibonacci sequence
    public static int fibonacciRec(int fibNumb) {


        if ((fibNumb == 1) || (fibNumb == 0)) {
            return fibNumb;



         }
        int answer = fibonacciRec(fibNumb - 1) + fibonacciRec(fibNumb - 2);
        return answer;
    }

    //Iteration method for the Fibonacci sequence
    static int fiboacciIte(int fibNumb) {
        int a = 0, b = 1, c = 1;
        for (int d = 0; d < fibNumb; d++) {
            a = b;
            b = c;
            System.out.println(a);
            c = a + b;
        }
        return a;   
    }   
}

Answer 1

你可以为递归函数添加一个单独的参数，它给出了到目前为止达到的最高Fibonacci下标（fibNumb）。如果当前下标大于此计数器，请更新它并打印当前的斐波纳契数。

计数器必须通过引用传递，因为在递归展开之前需要更新值。我们可以为它创建一个自定义类：

private static class Counter { private int value; public Counter() { value = 0; } public int getValue() { return value; } public void increment() { value++; } }

将相同的实例传递给每个递归调用并沿途递增：

private static int fibonacciRec_internal(int fibNumb_cur, Counter fibNumb_max) { int answer = (fibNumb_cur <= 1) ? fibNumb_cur : fibonacciRec_internal(fibNumb_cur - 1, fibNumb_max) + fibonacciRec_internal(fibNumb_cur - 2, fibNumb_max); if (fibNumb_max.getValue() < fibNumb_cur) { fibNumb_max.increment(); System.out.println(answer); } return answer; }

最后编写一个包装函数来隐藏这个计数器类及其用法细节：

public static int fibonacciRec(int fibNumb) { Counter max = new Counter(); return fibonacciRec_internal(fibNumb, max); }

结果如预期：

// fibonacciRec(10); 1 1 2 3 5 8 13 21 34 55

更好的是，我们可以为上面的所有创建一个包装类，并为每个fibonacciRec调用使用相同的对象，而不是创建一个新对象。这是一个非常微不足道的修改，所以我会留给你弄清楚它是什么以及如何。

递归和迭代fibonacci

1 个答案: