如何删除类之间的重复代码？

java dry

2022-09-03 15:03:14

我有2个类：递归斐波那契和记忆递归斐波那契。这就是我到目前为止所拥有的。

递归斐波那契类

public class SimpleRecursiveFibonacci {

  public BigInteger fibonacci(int n) {
    if(n < 2) {
      return BigInteger.ONE;             
    }

    return fibonacci(n - 2).add(fibonacci(n - 1));
  }
}

和记忆递归斐波那契类

public class MemoizedRecursiveFibonacci {
  private Map<Integer, BigInteger> cache = new HashMap<>();

  public BigInteger fibonacci(int n) {
    if(n < 2) {
      return BigInteger.ONE;
    }
    if(!cache.containsKey(n)){
      BigInteger currentFibonacci = fibonacci(n - 2).add(fibonacci(n - 1));
      cache.put(n, currentFibonacci);
    }

    return cache.get(n);
  }
}

正如我所看到的，在MemorizedRecursiveFibonacci类中有一些重复的代码

 if(n < 2) {
      return BigInteger.ONE;

和

  BigInteger currentFibonacci = fibonacci(n - 2).add(fibonacci(n - 1));

我怎样才能保持干燥？删除重复的代码？

答案 1

像这样的东西怎么样：

public class SimpleRecursiveFibonacci {

    /** Gets the fibonacci value for n */
    public final BigInteger fibonacci(int n) {
        if (n == 0) {
            return BigInteger.ZERO;
        } else if (n == 1) {
            return BigInteger.ONE;
        }
        return getFibonacci(n);
    }

    /** Recursively calculates the fibonacci by adding the two previous fibonacci. */
    protected final BigInteger calculateFibbonacci(int n) {
        return fibonacci(n - 2).add(fibonacci(n - 1));
    }

    /** 
     * Somehow get the fibonacci value for n.
     * Could be by calculation, getting it from a cache, or anything.
     * Defaults to calculation.
     */
    protected BigInteger getFibonacci(int n) {
        return calculateFibbonacci(n);
    }

}

public class MemoizedRecursiveFibonacci extends SimpleRecursiveFibonacci {

    // Cache using an array list as recommended by user @DodgyCodeException
    private ArrayList<BigInteger> cache = new ArrayList<>();

    @Override
    protected BigInteger getFibonacci(int n) {
        if (cache.size() < n) {
            BigInteger fib = calculateFibbonacci(n);
            cache.add(fib);
            return fib;
        } else {
            return cache.get(n - 1);
        }
    }
}

答案 2

也许这是一种选择...但我认为不是最好的。

public class SimpleRecursiveFibonacci {

public BigInteger fibonacci(int n) {
    if(n < 2) {
        return BigInteger.ONE;
    }
    return calculate(n);
}

protected BigInteger calculate(int n){
    return fibonacci(n - 2).add(fibonacci(n - 1)),
}

}

public class MemoizedRecursiveFibonacci extends SimpleRecursiveFibonacci{
private Map<Integer, BigInteger> cache = new HashMap<>();

@Override
protected BigInteger calculate(int n) {
    if(!cache.containsKey(n)){
        BigInteger currentFibonacci = super.calculate(n);
        cache.put(n, currentFibonacci);
    }
    return cache.get(n)
}

}