Reputation: 3386
How can we remove common values from two ArrayLists?
Let’s consider I have two Arraylist as shown below:
ArrayList1 = [1,2,3,4]
ArrayList1 = [2,3,4,6,7]
I would like to have result as:
ArrayListFinal = [1,6,7]
How can I do it?
Upvotes: 25
Views: 68933
Reputation: 672
You can use something like this:
ArrayList <Integer> first = new ArrayList <Integer> ();
ArrayList <Integer> second = new ArrayList <Integer> ();
ArrayList <Integer> finalResult = new ArrayList <Integer> ();
first.add(1);
first.add(2);
first.add(3);
first.add(4);
second.add(2);
second.add(3);
second.add(4);
second.add(6);
second.add(7);
for (int i = 0; i < first.size(); i++){
if (!second.contains(first.get(i))){
finalResult.add(first.get(i));
}
}
for (int j = 0; j < second.size(); j++){
if (!first.contains(second.get(j))){
finalResult.add(second.get(j));
}
}
I just populated two ArrayLists as you described them in your post, and I checked on both of them for distinct elements; if such element/s was/were found, I add them to the finalResult ArrayList.
Upvotes: 4
Reputation: 65811
You are actually asking for the Symmetric Difference.
List<Integer> aList = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
List<Integer> bList = new ArrayList<>(Arrays.asList(2, 3, 4, 6, 7));
// Union is all from both lists.
List<Integer> union = new ArrayList(aList);
union.addAll(bList);
// Intersection is only those in both.
List<Integer> intersection = new ArrayList(aList);
intersection.retainAll(bList);
// Symmetric difference is all except those in both.
List<Integer> symmetricDifference = new ArrayList(union);
symmetricDifference.removeAll(intersection);
System.out.println("aList: " + aList);
System.out.println("bList: " + bList);
System.out.println("union: " + union);
System.out.println("intersection: " + intersection);
System.out.println("**symmetricDifference: " + symmetricDifference+"**");
Prints:
aList: [1, 2, 3, 4]
bList: [2, 3, 4, 6, 7]
union: [1, 2, 3, 4, 2, 3, 4, 6, 7]
intersection: [2, 3, 4]
**symmetricDifference: [1, 6, 7]**
Upvotes: 22
Reputation: 726599
Here is an algorithm that you could follow to accomplish the task:
Java collections support addAll
, removeAll
, and retainAll
. Use addAll
to construct unions, retainAll
for constructing intersections, and removeAll
for subtraction, like this:
// Make the two lists
List<Integer> list1 = Arrays.asList(1, 2, 3, 4);
List<Integer> list2 = Arrays.asList(2, 3, 4, 6, 7);
// Prepare a union
List<Integer> union = new ArrayList<Integer>(list1);
union.addAll(list2);
// Prepare an intersection
List<Integer> intersection = new ArrayList<Integer>(list1);
intersection.retainAll(list2);
// Subtract the intersection from the union
union.removeAll(intersection);
// Print the result
for (Integer n : union) {
System.out.println(n);
}
Upvotes: 47
Reputation: 5940
SetList<Integer> A = new SetList<Integer>();
A.addAll({1,2,3,4});
SetList<Integer> B = new SetList<Integer>();
B.addAll({2,3,4,6,7});
Integer a = null;
for (int i=0; i<A.size(); i++)
{
a = A.get(i);
if (B.contains(a)
{
B.remove(a);
A.remove(a);
i--;
}
}
SetList<Integer> final = new SetList<Integer>();
final.addAll(A);
final.addAll(B);
// final = { 1, 6, 7 }
Upvotes: 0