java - java中的递归质因数算法

Question

我正在尝试在 java 中实现一个简单的算法来查找通过参数传递的整数的所有素数因子:

private static ArrayList<Integer> lstPrime= new ArrayList<Integer>();

    public static ArrayList primeFactors(int n) {

        if (isPrime(n)) 
        {
            lstPrime.add(n);
        }

        for (int i=2; i<=(Math.sqrt(n)); i++)
        {
            if (n % i == 0)
            {
                lstPrime.addAll(primeFactors(n/i));
                return lstPrime;
            }
        }
        return lstPrime;
    }

有趣的是，如果我将 81 作为 n 传递，结果将是:3, 3, 3, 3, 3, 3, 3, 3 而它应该是:3, 3, 3, 3 (3^4 =81)

Answer 1

可能有点复杂，但到目前为止它可以正常工作，并且它使用的可能是我能在 Java 中找到的最快(也是最小)的素数生成器。

首先，我们得到素数生成器，需要它来测试一个值是否为素数。我们使用生成器(这个比原始方法快 10 倍)所以我们使用缓存列表:

/**
 * Sieve of Sundaram prime number generator
 * Implementation following the Sieve of Sundaram to generate prime numbers 
 * 
 * @see http://en.wikipedia.org/wiki/Sieve_of_Sundaram
 */
static public class SundaramSievePrimeGenerator {
   public String getName() { return "Sieve of Sundaram generator"; }
   public int[] findPrimes(int max) {
      int n = max/2;
      boolean[] isPrime = new boolean[max];

      Arrays.fill(isPrime, true);

      for (int i=1; i<n; i++) {
         for (int j=i; j<=(n-i)/(2*i+1); j++) {
            isPrime[i+j+2*i*j] = false;
         }
      }

      int[] primes = new int[max];
      int found = 0;
      if (max > 2) {
         primes[found++] = 2;
      }
      for (int i=1; i<n; i++) {
         if (isPrime[i]) {
            primes[found++] = i*2+1;
         }
      }

      return Arrays.copyOf(primes, found);
   }
}

然后我们有了实际获取 n 的质因数列表所需的两种方法:

/**
 * Reuse an instance of the SundaramSievePrimeGenerator
 */
static public List<Integer> findPrimeFactors(int n, SundaramSievePrimeGenerator g) {
   ArrayList<Integer> primeFactors = new ArrayList<Integer>();

   int[] primes = g.findPrimes(n+1);
   int v;

   // debug
   //System.out.print("** primes found : ");
   //for (int a : primes) {
   //   System.out.print(" " + a);
   //}
   //System.out.println();

   if (primes[primes.length-1] == n) {
      primeFactors.add(n);
   } else {

      int max = primes.length - 1;

      for (int i=max; i>=0; i--) {
         primeFactors.add(primes[i]);
         if (testPrimeFactor(n, primes[i], primes, i, primeFactors)) {
            break;  // we found our solution
         }
         primeFactors.clear();
      }
   }

   return primeFactors;
}

/**
 * Recursive method initially called by findPrimeFactors(n, g)
 */
static private boolean testPrimeFactor(int n, int v, int[] primes, int index, List<Integer> factors) {
   int v2 = v * primes[index];

   if (v2 == n) {
      factors.add(primes[index]);
      return true;
   } else if (v2 > n) {
      if (index > 0) {
         return testPrimeFactor(n, v, primes, index-1, factors);
      } else {
         return false;
      }
   } else {
      while (index > 0) {
         factors.add(primes[index]);

         if (testPrimeFactor(n, v2, primes, index, factors)) {
            return true;
         }

         factors.remove(factors.size()-1);   // no good, remove added prime
         v2 = v * primes[--index];
      }
      return false;   // at this point, we are still below n... so no good
   }
}

最后，我们的测试用例:

int n = 1025;
SundaramSievePrimeGenerator generator = new SundaramSievePrimeGenerator();

List<Integer> factors = findPrimeFactors(n, generator);

if (factors.isEmpty()) {
   System.out.println("No prime factors found for " + n);
} else {
   System.out.println(n + " is composed of " + factors.size() + " prime factors");
   int v = 1;
   for (int i : factors) {
      v *= i;
      System.out.print(" " + i);
   }
   System.out.println(" = " + v);
}

例如，上面的这段代码将产生:

1025 is composed of 3 prime factors
 41 5 5 = 1025

更改n = 81 将产生所需的输出

81 is composed of 4 prime factors
 3 3 3 3 = 81