Another approach to find, if two lists have common elements is to use sets. The sets have unordered collection of unique elements. So we convert the lists into sets and then create a new set by combining the given sets. If they have some common elements then the new set will not be empty.
We can also apply the reduce function in python. This function is used to apply a given function passed onto it as argument to all of the list elements mentioned in the sequence passed along. The lambda function finds out the common elements by iterating through each nested list after set is applied to them .
Benefit:- If required, you can get the common elements as def commonItems = newList. intersect(oldList) . :) If you just need to check with a boolean you can also use ! newList.
Use Collection#retainAll()
.
listA.retainAll(listB);
// listA now contains only the elements which are also contained in listB.
If you want to avoid that changes are being affected in listA
, then you need to create a new one.
List<Integer> common = new ArrayList<Integer>(listA);
common.retainAll(listB);
// common now contains only the elements which are contained in listA and listB.
You can use set intersection operations with your ArrayList
objects.
Something like this:
List<Integer> l1 = new ArrayList<Integer>();
l1.add(1);
l1.add(2);
l1.add(3);
List<Integer> l2= new ArrayList<Integer>();
l2.add(4);
l2.add(2);
l2.add(3);
System.out.println("l1 == "+l1);
System.out.println("l2 == "+l2);
List<Integer> l3 = new ArrayList<Integer>(l2);
l3.retainAll(l1);
System.out.println("l3 == "+l3);
Now, l3
should have only common elements between l1
and l2
.
CONSOLE OUTPUT
l1 == [1, 2, 3]
l2 == [4, 2, 3]
l3 == [2, 3]
Why reinvent the wheel? Use Commons Collections:
CollectionUtils.intersection(java.util.Collection a, java.util.Collection b)
Using Java 8's Stream.filter()
method in combination with List.contains()
:
import static java.util.Arrays.asList;
import static java.util.stream.Collectors.toList;
/* ... */
List<Integer> list1 = asList(1, 2, 3, 4, 5);
List<Integer> list2 = asList(1, 3, 5, 7, 9);
List<Integer> common = list1.stream().filter(list2::contains).collect(toList());
consider two list L1 ans L2
Using Java8 we can easily find it out
L1.stream().filter(L2::contains).collect(Collectors.toList())
List<String> lista =new ArrayList<String>();
List<String> listb =new ArrayList<String>();
lista.add("Isabella");
lista.add("Angelina");
lista.add("Pille");
lista.add("Hazem");
listb.add("Isabella");
listb.add("Angelina");
listb.add("Bianca");
// Create an aplusb list which will contain both list (list1 and list2) in which common element will occur twice
List<String> listapluslistb =new ArrayList<String>(lista);
listapluslistb.addAll(listb);
// Create an aunionb set which will contain both list (list1 and list2) in which common element will occur once
Set<String> listaunionlistb =new HashSet<String>(lista);
listaunionlistb.addAll(listb);
for(String s:listaunionlistb)
{
listapluslistb.remove(s);
}
System.out.println(listapluslistb);
You can get the common elements between two lists using the method "retainAll". This method will remove all unmatched elements from the list to which it applies.
Ex.: list.retainAll(list1);
In this case from the list, all the elements which are not in list1 will be removed and only those will be remaining which are common between list and list1.
List<Integer> list = new ArrayList<>();
list.add(10);
list.add(13);
list.add(12);
list.add(11);
List<Integer> list1 = new ArrayList<>();
list1.add(10);
list1.add(113);
list1.add(112);
list1.add(111);
//before retainAll
System.out.println(list);
System.out.println(list1);
//applying retainAll on list
list.retainAll(list1);
//After retainAll
System.out.println("list::"+list);
System.out.println("list1::"+list1);
Output:
[10, 13, 12, 11]
[10, 113, 112, 111]
list::[10]
list1::[10, 113, 112, 111]
NOTE: After retainAll applied on the list, the list contains common element between list and list1.
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