Project Euler > Problem 50 > Consecutive prime sum (Java Solution)

Problem:

The prime 41, can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

Solution:

997651

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 50
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p050 implements EulerSolution {
 
 public static void main(String[] args) {
  System.out.println(new p050().run());
 }
 
 
 private static final int LIMIT = Library.pow(10, 6);
 
 
 public String run() {
  boolean[] isPrime = Library.listPrimality(LIMIT);
  int[] primes = Library.listPrimes(LIMIT);
  
  long maxSum = 0;
  int maxRun = -1;
  for (int i = 0; i < primes.length; i++) {  // For each index of a starting prime number
   int sum = 0;
   for (int j = i; j < primes.length; j++) {  // For each end index (inclusive)
    sum += primes[j];
    if (sum > LIMIT)
     break;
    else if (j - i > maxRun && sum > maxSum && isPrime[sum]) {
     maxSum = sum;
     maxRun = j - i;
    }
   }
  }
  return Long.toString(maxSum);
 }
 
}