Is there any way to determine whether an ArrayList contains any element of a different ArrayList?
Like this:
list1.contains(any element of list2)
Is looping through all the elements of list2 and checking the elements one by one the only way?
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If 'X contains an element of Y' is the main usecase for your collection, you may want to consider using Set instead.maasg– maasg2013-09-22 12:41:14 +00:00Commented Sep 22, 2013 at 12:41
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6 Answers
Consider the following: Java SE 7 documentation: java.util.Collections.disjoint
The "disjoint" method takes two collections (listA and listB for example) as parameters and returns "true" if they have no elements in common; thus, if they have any elements in common, it will return false.
A simple check like this is all that's required:
if (!Collections.disjoint(listA, listB))
{
//List "listA" contains elements included in list "listB"
}
2 Comments
ArifMustafa
very useful method 1.7 onwards... :D
Ernani
I prefer this answer as it uses native Java methods only and there is no casting required.
Although not highly efficient, this is terse and employs the API:
if (!new HashSet<T>(list1).retainAll(list2).isEmpty())
// at least one element is shared
Comments
if(!CollectionUtils.intersection(arrayList1, arrayList2).isEmpty()){
// has common
}
else{
//no common
}
use org.apache.commons.collections
Comments
If you have access to Apache Commons, see CollectionUtils.intersection(a,b)
Use like this:
! CollectionUtils.intersection(list1, list2).isEmpty()
Comments
How about trying like this:-
List1.retainAll(List2)
like this:-
int a[] = {30, 100, 40, 20, 80};
int b[] = {100, 40, 120, 30, 230, 10, 80};
List<Integer> 1ist1= Arrays.asList(a);
List<Integer> 1ist2= Arrays.asList(b);
1ist1.retainsAll(1ist2);
Comments
If you're not constrained in using third-party libraries, Apache commons ListUtils is good for common list operations.
In this case you could use the intersection method
if(!ListUtils.intersection(list1,list2).isEmpty()) {
// list1 & list2 have at least one element in common
}
