斐波那契递归不适用于 Java答案

【问题标题】：Fibonacci Recursion not working for Java斐波那契递归不适用于 Java
【发布时间】：2017-06-27 06:12:32
【问题描述】：

请原谅我的无知，我还是个菜鸟。每当我输入一个中等大的数字时，它不会给我任何输出，因为它卡在 fibCalculate 方法中。不提供任何输出的数字示例是 40。

package fib;

import java.math.BigInteger;
import java.util.Scanner;


public class FibonacciCalculator {
    private static BigInteger TWO = BigInteger.valueOf(2);

    public static BigInteger fibCalculate(BigInteger num) {
        if (num.equals(BigInteger.ONE) || num.equals(BigInteger.ZERO)) {
            return num;
        } else {
            return (fibCalculate(num.subtract(BigInteger.ONE)).add(fibCalculate(num.subtract(TWO))));
        }
    }

    public static void main(String[] args) {
        @SuppressWarnings("resource")
        Scanner input = new Scanner(System.in);
        long n = 10;

        System.out.println(
                "Please input a fibonacci placeholder to see what fibonacci number it holds,\nor input 0 to end program. (The count starts at 0):  ");
        n = input.nextLong();
        System.out.println("Answer: "+fibCalculate(BigInteger.valueOf(n)));

    }
}

【问题讨论】：

这通常是斐波那契预期的结果。为什么不能使用迭代方法？
我很惊讶你能走到这一步。您使用的方法非常慢。仅使用一个循环搜索替代品（网络上到处都是）。
您的代码确实会产生 n>40 的输出，只是需要很多时间，因为它非常慢。
您的算法与n 呈指数关系。每次计算斐波那契数时，您都在计算两个较小的斐波那契数，依此类推。您应该尝试设计另一种算法。改进这一点的一种方法是将结果存储在某种数组中，并尝试重复使用它们。

标签： java recursion fibonacci

【解决方案1】：

您的代码确实会产生 n>40 的输出，但这需要很多时间，因为它非常慢。

例如，使用以下代码测试代码中所有介于 0 和 50 之间的 n 值：

long start = System.currentTimeMillis ();
for (int fib=0;fib<50;fib++) {
  System.out.println("Answer: "+ fib + ": "+fibCalculate(BigInteger.valueOf(fib)) + " after " + (System.currentTimeMillis ()-start)/1000 + " seconds from the beginning");
}

产生以下输出（我没有等到它结束）：

Answer: 0: 0 after 0 seconds from the beginning
Answer: 1: 1 after 0 seconds from the beginning
Answer: 2: 1 after 0 seconds from the beginning
Answer: 3: 2 after 0 seconds from the beginning
Answer: 4: 3 after 0 seconds from the beginning
Answer: 5: 5 after 0 seconds from the beginning
Answer: 6: 8 after 0 seconds from the beginning
Answer: 7: 13 after 0 seconds from the beginning
Answer: 8: 21 after 0 seconds from the beginning
Answer: 9: 34 after 0 seconds from the beginning
Answer: 10: 55 after 0 seconds from the beginning
Answer: 11: 89 after 0 seconds from the beginning
Answer: 12: 144 after 0 seconds from the beginning
Answer: 13: 233 after 0 seconds from the beginning
Answer: 14: 377 after 0 seconds from the beginning
Answer: 15: 610 after 0 seconds from the beginning
Answer: 16: 987 after 0 seconds from the beginning
Answer: 17: 1597 after 0 seconds from the beginning
Answer: 18: 2584 after 0 seconds from the beginning
Answer: 19: 4181 after 0 seconds from the beginning
Answer: 20: 6765 after 0 seconds from the beginning
Answer: 21: 10946 after 0 seconds from the beginning
Answer: 22: 17711 after 0 seconds from the beginning
Answer: 23: 28657 after 0 seconds from the beginning
Answer: 24: 46368 after 0 seconds from the beginning
Answer: 25: 75025 after 0 seconds from the beginning
Answer: 26: 121393 after 0 seconds from the beginning
Answer: 27: 196418 after 0 seconds from the beginning
Answer: 28: 317811 after 0 seconds from the beginning
Answer: 29: 514229 after 0 seconds from the beginning
Answer: 30: 832040 after 0 seconds from the beginning
Answer: 31: 1346269 after 0 seconds from the beginning
Answer: 32: 2178309 after 0 seconds from the beginning
Answer: 33: 3524578 after 1 seconds from the beginning
Answer: 34: 5702887 after 1 seconds from the beginning
Answer: 35: 9227465 after 2 seconds from the beginning
Answer: 36: 14930352 after 4 seconds from the beginning
Answer: 37: 24157817 after 6 seconds from the beginning
Answer: 38: 39088169 after 10 seconds from the beginning
Answer: 39: 63245986 after 17 seconds from the beginning
Answer: 40: 102334155 after 28 seconds from the beginning
Answer: 41: 165580141 after 45 seconds from the beginning
Answer: 42: 267914296 after 74 seconds from the beginning
Answer: 43: 433494437 after 120 seconds from the beginning

可以看到运行时间呈指数增长。

【讨论】：

【解决方案2】：

您实现的计算将如下所示：

                          fibCalculate(40)
                         /                \
                fibCalculate(39)       fibCalculate(38)   
               /            \           /             \
fibCalculate(38)  fibCalculate(37)  fibCalculate(37)   fibCalculate(36)
     /     \           /     \         /     \            /     \
 ....     ....      ....   ....      ....   ....       ....    ....

以此类推，直到它达到fibCalculate(0) 或fibCalculate(1)，导致总体上进行 2^40（O(2^n) 运行时间复杂度）计算，其中大部分是相同计算值的重复，因此几乎立即的优化将缓存中间值以避免第二次计算它们，这会将运行时间复杂度降低到 O(n)。

package fib;

import java.math.BigInteger;
import java.util.Scanner;


public class FibonacciCalculator {
   private static Map<BigInteger, BigInteger> cache = new HashMap<>() {{
       put(BigInteger.ONE);
       put(BigInteger.valueOf(2));
   }}

public static BigInteger fibCalculate(BigInteger num) {
       BigInteger result = cache.get(num);
       if (result != null) {
           return result
       }
       result = (fibCalculate(num.subtract(BigInteger.ONE)).add(fibCalculate(num.subtract(TWO))));
       return result;
    }
}

public static void main(String[] args) {
    @SuppressWarnings("resource")
    Scanner input = new Scanner(System.in);
    long n = 10;

    System.out.println(
            "Please input a fibonacci placeholder to see what fibonacci number it holds,\nor input 0 to end program. (The count starts at 0):  ");
    n = input.nextLong();
    System.out.println("Answer: "+fibCalculate(BigInteger.valueOf(n)));

   }
}

【讨论】：