Method factorial

output/java_guava/1.4.19/BigIntegerMath.java:339–392

Returns n!, that is, the product of the first n positive integers, or 1 if n == 0. Warning: the result takes O(n log n) space, so use cautiously. This uses an efficient binary recursive algorithm to compute the factorial with balanced multiplies.

(int n)

  public static BigInteger factorial(int n) {
    checkNonNegative("n", n);

    // If the factorial is small enough, just use LongMath to do it.
    if (n < LongMath.factorials.length) {
      return BigInteger.valueOf(LongMath.factorials[n]);
    }

    // Pre-allocate space for our list of intermediate BigIntegers.
    int approxSize = IntMath.divide(n * IntMath.log2(n, CEILING), Long.SIZE, CEILING);
    ArrayList<BigInteger> bignums = new ArrayList<BigInteger>(approxSize);

    // Start from the pre-computed maximum long factorial.
    int startingNumber = LongMath.factorials.length;
    long product = LongMath.factorials[startingNumber - 1];
    // Strip off 2s from this value.
    int shift = Long.numberOfTrailingZeros(product);
    product >>= shift;

    // Use floor(log2(num)) + 1 to prevent overflow of multiplication.
    int productBits = LongMath.log2(product, FLOOR) + 1;
    int bits = LongMath.log2(startingNumber, FLOOR) + 1;
    // Check for the next power of two boundary, to save us a CLZ operation.
    int nextPowerOfTwo = 1 << (bits - 1);

    // Iteratively multiply the longs as big as they can go.
    for (long num = startingNumber; num <= n; num++) {
      // Check to see if the floor(log2(num)) + 1 has changed.
      if ((num & nextPowerOfTwo) != 0) {
        nextPowerOfTwo <<= 1;
        bits++;
      }
      // Get rid of the 2s in num.
      int tz = Long.numberOfTrailingZeros(num);
      long normalizedNum = num >> tz;
      shift += tz;
      // Adjust floor(log2(num)) + 1.
      int normalizedBits = bits - tz;
      // If it won't fit in a long, then we store off the intermediate product.
      if (normalizedBits + productBits >= Long.SIZE) {
        bignums.add(BigInteger.valueOf(product));
        product = 1;
        productBits = 0;
      }
      product *= normalizedNum;
      productBits = LongMath.log2(product, FLOOR) + 1;
    }
    // Check for leftovers.
    if (product > 1) {
      bignums.add(BigInteger.valueOf(product));
    }
    // Efficiently multiply all the intermediate products together.
    return listProduct(bignums).shiftLeft(shift);
  }


  static BigInteger listProduct(List<BigInteger> nums) {