How to construct ArrayList from 2D int array?

If you want all unique numbers straight away then you could use Set<Integer> directly

30,40,50
50,30,40
30,40,50

in above example 2nd & 3rd row is duplicate of 1st row

if you see all 3 rows same then You could directly use Set<Integer>

First, you should create a Class to hold you data which override public boolean equals(Object obj) and public int hashCode() to indicate equality of the data.

public class Row {

    private int[] ints;

    public Row(int[] ints) {
        this.ints = ints.clone();
        Arrays.sort(this.ints);
    }

    @Override
    public int hashCode() {
        return Arrays.hashCode(ints);
    }

    @Override
    public boolean equals(Object obj) {

        if(obj instanceof Row) {
            Row another = (Row) obj;
            int[] original = Arrays.copyOf(another.ints, another.ints.length);
            return Arrays.equals(ints, original);
        } else {
            return false;
        }
    }

    @Override
    public String toString() {
        return Arrays.toString(ints);
    }

}

Test Case

public class Test {

    public static void main(String[] args) {
        int[][] arrays = new int[][]{{30,40,50}, {50,30,40}, {30,40,50}, {10, 20, 30}};
        Set<Row> rows = new HashSet<Row>();
        for(int[] a: arrays) {
            rows.add(new Row(a));
        }
        for(Row row: rows) {
            System.out.println(row);
        }
    }
}

Output

[10, 20, 30]
[30, 40, 50]

// lets create table
int[][] int2d = { 
        { 1, 2 }, 
        { 2, 1 }, 
        { 30, 40, 50 }, 
        { 50, 30, 40 },
        { 30, 40, 50 }, };

// lets sort content of each row for
for (int[] row : int2d)
    Arrays.sort(row);
// lets see how table looks likne now
System.out.println(Arrays.deepToString(int2d));

// this will help set do decide if element is already in
Comparator<int[]> comparator = new Comparator<int[]>() {
    public int compare(int[] o1, int[] o2) {
        return Arrays.toString(o1).compareTo(Arrays.toString(o2));
    }
};

// we create set and give him comparator (via constructor)
Set<int[]> set = new TreeSet<int[]>(comparator);
// now lets try to put every row of table in set
for (int[] row : int2d)
    set.add(row);

// NOW, GREAT TEST
System.out.println("============");
for (int[] row : set)
    System.out.println(Arrays.toString(row));

out:

[[1, 2], [1, 2], [30, 40, 50], [30, 40, 50], [30, 40, 50]]
============
[1, 2]
[30, 40, 50]

This is a minimal code that will eliminate the duplicates as you have defined them:

final Integer[][] int2d = { { 30, 40, 50 }, { 50, 30, 40 }, { 30, 40, 50 }, };
final Set<Set<Integer>> r = new LinkedHashSet<Set<Integer>>();
for (Integer[] row : int2d) r.add(new LinkedHashSet<Integer>(Arrays.asList(row)));
System.out.println(r);

If you use the persistent collections from Functional Java, the solution is quite simple.

(Note: List and Set I am referring to below are from Functional Java, and not from standard library.)

Let us say you are storing your data in a variable named xss of type List<List<Integer>>.

What you want to do can be described as follows:

Deduplicate xss with the criterion that the two rows that contain same elements are to be considered equal.

• List provides a method nub for deduplication.
• Whether or not the two Lists contain same elements can be found out by converting them to Sets and then comparing them for equality.
• The above criterion can be passed as a Equal instance.

Code:

xss.nub(new Equal<List<Integer>>() {
  public boolean eq(List<Integer> xs1, List<Integer> xs2) {
    Set<Integer> s1 = Set.iterableSet(Ord.listOrd(Ord.intOrd), xs1);
    Set<Integer> s2 = Set.iterableSet(Ord.listord(Ord.intOrd), xs2);
    return Equal.setEqual(Equal.intEqual).eq(s1, s2);
  }
});