web development helpdesk

    public static <T> T[] arrayMerge(T[]... arrays)
    {
        // Determine required size of new array

        int count = 0;
        for (T[] array : arrays)
        {
            count += array.length;
        }

        // create new array of required class

        T[] mergedArray = (T[]) Array.newInstance(
           arrays[0][0].getClass(),count);

        // Merge each array into new array

        int start = 0;
        for (T[] array : arrays)
        {
            System.arraycopy(array, 0, 
               mergedArray, start, array.length);
            start += array.length;
        }
        return (T[]) mergedArray;
    } 

11 Responses to “How can I merge two Java arrays into one combined array?”

1. Kamran Says:
   May 5th, 2009 at 4:12 pm

   Here is another variant…

   public static T[] merge(T[]… arrays) {
   List list = new ArrayList();

   for( T[] array : arrays )
   list.addAll( Arrays.asList( array ) );

   return (T[]) Array.newInstance( arrays[0][0].getClass(), list.size() );
   }

2. Rarish Ravi Says:
   May 18th, 2010 at 11:20 am

   public static void merge(String[] a, String[] b)
   {
   List lst = new ArrayList(Arrays.asList(a));
   lst.addAll(Arrays.asList(b));

   Object[] result = lst.toArray();

   System.out.println(“Merged output result is “+Arrays.toString(result));
   }

3. abc Says:
   December 28th, 2010 at 1:31 am

   hi

4. def Says:
   April 8th, 2011 at 4:01 am

   yo

5. Ganesh Says:
   August 8th, 2011 at 6:28 pm

   public class array1 {
   public static void merge(String[] a, String[] b)
   {
   List lst = new ArrayList(Arrays.asList(a));
   lst.addAll(Arrays.asList(b));

   Object[] result = lst.toArray();
   Arrays.sort(result);

   System.out.println(“Merged output result is”+Arrays.toString(result));
   }

   public static void main(String arg[]){
   String a[]={“1″,”2″,”3″,”4″,”5″};
   String b[]={“2″,”3″,”1″,”4″,”5″};
   array1.merge(a, b);

   }
   }

6. Emanuele Says:
   November 10th, 2011 at 10:48 pm

   All the variants posted here in the comments are pointless, unefficient and, most of all, get rid of the extremely efficient System.arraycopy.

7. Fmanuele Says:
   November 15th, 2011 at 10:56 pm

   @Emanuele why dont you tell us your genius-blessed own methods?

8. Emanuele Says:
   December 12th, 2011 at 11:41 pm

   @Fmanuele: my favourite version is the one posted into the article. I was going to design my own implementation using System.arraycopy, in a similar fashion to this article, when I came across to it. I just simply adapted to my purposes, since I found it very clever.

9. Nicolas Dos Santos Says:
   March 28th, 2012 at 11:49 pm

   This versions require Java 6, as they use Arrays.copyOf()

   public static T[] concat(T[] first, T[]… rest) {
   int totalLength = first.length;
   for (T[] array : rest) {
   totalLength += array.length;
   }
   T[] result = Arrays.copyOf(first, totalLength);
   int offset = first.length;
   for (T[] array : rest) {
   System.arraycopy(array, 0, result, offset, array.length);
   offset += array.length;
   }
   return result;
   }

10. austin Says:
    May 29th, 2012 at 6:04 pm

    Thanks! =)

11. rohit kasat Says:
    August 3rd, 2012 at 1:25 pm

    for o(n) complexity 3 array merge

    public static ArrayList merge(int al[], int bl[], int cl[]) {
    ArrayList tl = new ArrayList();
    int ai, bi, ci;
    ai = bi = ci = 0;

    while (true) {
    if (ai != al.length) {
    if (bi != bl.length) {
    if (al[ai] < bl[bi]) {
    if (ci != cl.length) {
    if (al[ai] < cl[ci]) {
    tl.add(al[ai]);
    ai++;
    } else {
    tl.add(cl[ci]);
    ci++;
    }
    } else {
    tl.add(al[ai]);
    ai++;
    }
    } else {
    if (ci != cl.length) {
    if (bl[bi] < cl[ci]) {
    tl.add(bl[bi]);
    bi++;
    } else {
    tl.add(cl[ci]);
    ci++;
    }
    } else {
    tl.add(bl[bi]);
    bi++;
    }

    }
    } else {
    if (ci != cl.length) {
    if (al[ai] < cl[ci]) {
    tl.add(al[ai]);
    ai++;
    } else {
    tl.add(cl[ci]);
    ci++;
    }
    } else {
    tl.add(al[ai]);
    ai++;
    }
    }
    } else if (bi != bl.length) {
    if (ci != cl.length) {
    if (bl[bi] < cl[ci]) {
    tl.add(bl[bi]);
    bi++;
    } else {
    tl.add(cl[ci]);
    ci++;
    }
    } else {
    tl.add(bl[bi]);
    bi++;
    }

    } else if (ci != cl.length) {
    tl.add(cl[ci]);
    ci++;
    } else {
    break;
    }
    }
    return tl;
    }

Leave a Reply