java - 将几种方法组合成唯一的一种。泛型

Question

我有一些代码可以确定与 BigInteger 相比，对不同数据类型(int、long、double)的操作的正确性。操作是获取数字的阶乘，直到结果与 BigInteger 相同。

问题是我如何更改我的代码，使其更通用、紧凑和干净？我怎么能得到唯一的一种方法，而不是 4 种不同类型的方法？ 此方法中的逻辑和流程几乎相同。

代码(没有比较逻辑)是:

private static HashMap<BigInteger, BigInteger> bigIntegerFactorials = new HashMap<>();
private static BigInteger bigIntegerFactorial(BigInteger number) {
    if (number.equals(BigInteger.ONE)) {
        return BigInteger.ONE;
    }
    BigInteger result = bigIntegerFactorials.get(number);
    if (result == null) {
        result = number.multiply(bigIntegerFactorial(number.subtract(BigInteger.ONE)));
        bigIntegerFactorials.put(number, result);
    }
    return result;
}

private static HashMap<Integer, Integer> intFactorials = new HashMap<>();
private static int intFactorial(int number) {
    if (number == 1) {
        return 1;
    }
    Integer result = intFactorials.get(number);
    if (result == null) {
        result = number * intFactorial(number - 1);
        intFactorials.put(number, result);
    }
    return result;
}

private static HashMap<Long, Long> longFactorials = new HashMap<>();
private static long longFactorial(long number) {
    if (number == 1) {
        return 1L;
    }
    Long result = longFactorials.get(number);
    if (result == null) {
        result = number * longFactorial(number - 1);
        longFactorials.put(number, result);
    }
    return result;
}

private static HashMap<Double, Double> doubleFactorials = new HashMap<>();
private static double doubleFactorial(double number) {
    if (number == 1) {
        return 1.;
    }
    Double result = doubleFactorials.get(number);
    if (result == null) {
        result = number * doubleFactorial(number - 1);
        doubleFactorials.put(number, result);
    }
    return result;
}

非常感谢。

Answer 1

您可以将乘法和减量函数传递给通用方法:

private static Map<Number, Number> factorials = new HashMap<> ();

private static <T extends Number> T factorial(T n, BinaryOperator<T> multiply, UnaryOperator<T> decrement) {
  if (n.doubleValue() == 1) return n;
  T result = (T) factorials.get(n);
  if (result == null ){
    result = multiply.apply(n, factorial(decrement.apply(n), multiply, decrement));
    factorials.put(n, result);
  }
  return result;
}

然后你可以像这样改变你的原始方法:

public static int intFactorial(int number) {
  return factorial(number, (i, j) -> i * j, i -> i - 1);
}

警告:此方法似乎会使 Netbeans 崩溃，但可以使用 javac 正常编译...