Alex Walker

Of course, I see it now. Thanks for clearing that up.

Alex

I realize that I'm kind of bringing this thread back from the dead, but I ran across this problem myself and I'm really trying to understand how best to implement the HashTable solution and why it's more efficient than the brute force approach. I came up with the class below, but I'm still a bit unclear on how this is more efficient.

Thanks,
Alex

import java.util.HashSet; import java.util.Set; public class IntIntersection { public static void main(String[] args) { int[] firstInts = new int[]{2,3,4,5,5,7,9,0,0,0}; int[] secondInts = new int[]{2,44,2,33,8,5,12,14,0}; Set<Integer> intersects = intersectArrays(firstInts, secondInts); // print out the set of intersecting intergers for (Integer theInt : intersects) { System.out.println("intersect on " + theInt); } } public static Set<Integer> intersectArrays(int[] first, int[] second) { // initialize a return set for intersections Set<Integer> intsIntersect = new HashSet<Integer>(); // load first array to a hash HashSet<Integer> array1ToHash = new HashSet<Integer>(); for (int i = 0; i < first.length; i++) { array1ToHash.add(first[i]); } // check second array for matches within the hash for (int i = 0; i < second.length; i++) { if (array1ToHash.contains(second[i])) { // add to the intersect array intsIntersect.add(second[i]); } } return intsIntersect; } }