array_intersect in java

Question

I found this snippet online that made a array_intersect like function in java using integer.

Sample:

        int[] intersect(int[] arr1, int[] arr2) {
            int count = 0;
            for(int a = 0; a < arr1.length; a++) {
                for(int b = 0; b < arr2.length; b++) {
                    if(arr1[a] == arr2[b]) {
                        count++;
                        break;
                    }
                }
            }

            int[] result = new int[count];
            count = 0;
            for(int a = 0; a < arr1.length; a++) {
                for(int b = 0; b < arr2.length; b++) {
                    if(arr1[a] == arr2[b]) {
                        result[count++] = arr1[a];
                        break;
                    }
                }
            }

            return result;
        }

        int[] arr1 = new int[] {10, 20, 30, 40, 50, 60, 70, 80, 90, 100,
         95, 85, 75, 65, 55, 45, 35, 25, 15, 05,
         10, 15, 20, 25, 30, 35, 40, 45, 50, 55};

        int[] arr2 = new int[] {15, 25, 35, 45, 55,
         12, 22, 32, 43, 52,
         15, 25, 35, 45, 55};

        int[] p1 = this.unique(arr1);
        int[] p2 = this.unique(arr2);
        int[] intersectResults = this.intersect(arr1, arr2);

            for(int a = 0; a < intersectResults.length; a++) {
                System.out.print(intersectResults[a] + "    ");
            }

But when i changed it to:

        String[] intersect(String[] a_yourname, String[] a_crushname) {
            int count = 0;
            for(int a = 0; a < a_yourname.length; a++) {
                for(int b = 0; b < a_crushname.length; b++) {
                    if(a_yourname[a] == a_crushname[b]) {
                        count++;
                        break;
                    }
                }
            }

            String[] result = new String[count];
            count = 0;
            for(int a = 0; a < a_yourname.length; a++) {
                for(int b = 0; b < a_crushname.length; b++) {
                    if(a_yourname[a] == a_crushname[b]) {
                        result[count++] = a_yourname[a];
                        break;
                    }
                }
            }

            return result;
        }

     String[] flames = this.intersect(a_yourname, a_crushname);
            //String[] p23 = this.unique(arr2);
             System.out.println("heheh" +flames.length);
             for(int a = 0; a < flames.length; a++) {
                System.out.print(flames[a] + "    ");
            }

Can someone explain to me what I did wrong here? I'm really not familiar with Java.

a_yourname and a_crushname are both string arrays.

Are you using == to compare two strings? The cleaner and potentially faster method is to put one array into set and another into some collection (e.g. Arrays.asList()) and call retainAll() on the set. — billc.cn
– billc.cn, Commented Jan 26, 2013 at 10:59
@billc.cn can you give an example of that? i just got that code form the net not sure how that works :) — magicianiam
– magicianiam, Commented Jan 26, 2013 at 11:27
Putting Java code incorrectly in a JSP file instead of a Java class doesn't make it a JSP problem. You'd have had exactly the same problem when putting that Java code in a normal Java class with a main() method. I removed the non-related JavaEE/JSP noise from the question. — BalusC
– BalusC, Commented Jan 26, 2013 at 12:40

billc.cn · Accepted Answer · 2013-01-26 14:53:13Z

1

Code:

    String[] a = new String[] { "aa", "bb", "cc" };
    String[] b = new String[] { "bb", "cc", "dd" };

    Set<String> setA = new HashSet<String>(Arrays.asList(a));
    List<String> listB = Arrays.asList(b);
    setA.retainAll(listB);
    System.out.println(setA);

You can see the results here: http://ideone.com/tKg5xa

If you really want the result as an array:

String[] out = setA.toArray(new String[setA.size()]);

Again there's no need to use an array if all you want is to iterate through the result. Google "Java for each" or pick up a Java book.

edited Jan 26, 2013 at 14:53

answered Jan 26, 2013 at 11:32

billc.cn

7,3443 gold badges43 silver badges82 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

magicianiam Over a year ago

Why is there a new String[]? Is it necessary or is just a choice of declaration of array?

billc.cn Over a year ago

No, it's not strictly necessary in this case. It's just an old habit.

magicianiam Over a year ago

is setA returned as an Array?

billc.cn Over a year ago

No, it's a Set. Use Set.toArray() to get an array, but there's really no need to do that. Most cases Collections works better than Arrays in modern Java.

magicianiam Over a year ago

thank you for this it really works great, but regarding the code above do you understand most of it?

|

Collectives™ on Stack Overflow

array_intersect in java

1 Answer 1

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related