How to calculate 2 to-the-power N where N is a very large number

User_67128 :

I need to find 2 to-the-power N where N is a very large number (Java BigInteger type)

Java BigInteger Class has pow method but it takes only integer value as exponent.

So, I wrote a method as follows:

static BigInteger twoToThePower(BigInteger n)
   {
      BigInteger result = BigInteger.valueOf(1L);

      while (n.compareTo(BigInteger.valueOf((long) Integer.MAX_VALUE)) > 0)
      {
         result = result.shiftLeft(Integer.MAX_VALUE);
         n = n.subtract(BigInteger.valueOf((long) Integer.MAX_VALUE));

      }

      long k = n.longValue();
      result = result.shiftLeft((int) k);

      return result;
   }

My code works fine, I am just sharing my idea and curious to know if there is any other better idea?

Thank you.

Esc Điệp :

Emmanuel Lonca's answer is correct. But, by Manoj Banik's idea, I would like to share my idea too.

My code do the same thing as Manoj Banik's code in faster way. The idea is init the buffer, and put the bit 1 in to correct location. I using the shift left operator on 1 byte instead of shiftLeft method.
Here is my code:

    static BigInteger twoToThePower(BigInteger n){
        BigInteger eight = BigInteger.valueOf(8);
        BigInteger[] devideResult = n.divideAndRemainder(eight);
        BigInteger bufferSize = devideResult[0].add(BigInteger.ONE);
        int  offset = devideResult[1].intValue();
        byte[] buffer = new byte[bufferSize.intValueExact()];
        buffer[0] = (byte)(1 << offset);
        return new BigInteger(1,buffer);
    }

But it still slower than BigInteger.pow

Then, I found that class BigInteger has a method called setBit. It also accepts parameter type int like the pow method. Using this method is faster than BigInteger.pow.
The code can be:

    static BigInteger twoToThePower(BigInteger n){
        return BigInteger.ZERO.setBit(n.intValueExact());
    }

Class BigInteger has a method called modPow also. But It need one more parameter. This means you should specify the modulus and your result should be smaller than this modulus. I did not do a performance test for modPow, but I think it should slower than the pow method.