find nth prime number

Question

I've written the following code below to find the nth prime number. Can this be improved in time complexity?

Description:

The ArrayList arr stores the computed prime numbers. Once arr reaches a size 'n', the loop exits and we retrieve the nth element in the ArrayList. Numbers 2 and 3 are added before the prime numbers are calculated, and each number starting from 4 is checked to be prime or not.

public void calcPrime(int inp) {
    ArrayList<Integer> arr = new ArrayList<Integer>(); // stores prime numbers 
                                                      // calculated so far
    // add prime numbers 2 and 3 to prime array 'arr'
    arr.add(2); 
    arr.add(3);

    // check if number is prime starting from 4
    int counter = 4;
     // check if arr's size has reached inp which is 'n', if so terminate while loop
    while(arr.size() <= inp) {
        // dont check for prime if number is divisible by 2
        if(counter % 2 != 0) {
            // check if current number 'counter' is perfectly divisible from 
           // counter/2 to 3
            int temp = counter/2;
            while(temp >=3) {
                if(counter % temp == 0)
                    break;
                temp --;
            }
            if(temp <= 3) {
                arr.add(counter);
            }
        }
        counter++;
    }

    System.out.println("finish" +arr.get(inp));
    }
}

Answer 1

Yes.

Your algorithm make O(n^2) operations (maybe I'm not accurate, but seems so), where n is result.

There are http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes algorithm that takes O(ipn* log(log(n))). You can make only inp steps in it, and assume that n = 2ipn*ln(ipn). n just should be greater then ipn-prime. (we know distributions of prime numbers http://en.wikipedia.org/wiki/Prime_number_theorem)

Anyway, you can improve existing solution:

public void calcPrime(int inp) {
    ArrayList<Integer> arr = new ArrayList<Integer>();
    arr.add(2);
    arr.add(3);

    int counter = 4;

    while(arr.size() < inp) {
        if(counter % 2 != 0 && counter%3 != 0) {
            int temp = 4;
            while(temp*temp <= counter) {
                if(counter % temp == 0)
                    break;
                temp ++;
            }
            if(temp*temp > counter) {
                arr.add(counter);
            }
        }
        counter++;
    }

    System.out.println("finish" +arr.get(inp-1));
    }
}