Project Euler > Problem 32 > Pandigital products (Java Solution)

Problem:

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 [×] 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

Solution:

45228

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 32
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */

import java.util.Arrays;


public final class p032 implements EulerSolution {
 
 public static void main(String[] args) {
  System.out.println(new p032().run());
 }
 
 
 public String run() {
  // A candidate product has at most 4 digits. This is because if it has 5 digits,
  // then the two multiplicands must have at least 5 digits put together.
  int sum = 0;
  for (int i = 1; i < 10000; i++) {
   if (hasPandigitalProduct(i))
    sum += i;
  }
  return Integer.toString(sum);
 }
 
 
 private static boolean hasPandigitalProduct(int n) {
  // Find and examine all factors of n
  for (int i = 1; i <= n; i++) {
   if (n % i == 0 && isPandigital("" + n + i + n/i))
    return true;
  }
  return false;
 }
 
 
 private static boolean isPandigital(String s) {
  if (s.length() != 9)
   return false;
  char[] temp = s.toCharArray();
  Arrays.sort(temp);
  return new String(temp).equals("123456789");
 }
 
}