Is there a better way to combine two string sets in java?

I need to combine two string sets while filtering out redundant information, this is the solution I came up with, is there a better way that anyone can suggest? Perhaps something built in that I overlooked? Didn't have any luck with google.

Set<String> oldStringSet = getOldStringSet();
Set<String> newStringSet = getNewStringSet();
        
for(String currentString : oldStringSet)
{
    if (!newStringSet.contains(currentString))
    {
        newStringSet.add(currentString);
    }
}

Since a Set does not contain duplicate entries, you can therefore combine the two by:

newStringSet.addAll(oldStringSet);

It does not matter if you add things twice, the set will only contain the element once... e.g it's no need to check using contains method.

You can do it using this one-liner

Set<String> combined = Stream.concat(newStringSet.stream(), oldStringSet.stream())
        .collect(Collectors.toSet());

With a static import it looks even nicer

Set<String> combined = concat(newStringSet.stream(), oldStringSet.stream())
        .collect(toSet());

Another way is to use flatMap method:

Set<String> combined = Stream.of(newStringSet, oldStringSet).flatMap(Set::stream)
        .collect(toSet());

Also any collection could easily be combined with a single element

Set<String> combined = concat(newStringSet.stream(), Stream.of(singleValue))
        .collect(toSet());

The same with Guava:

Set<String> combinedSet = Sets.union(oldStringSet, newStringSet)

From the definition Set contain only unique elements.

Set<String> distinct = new HashSet<String>(); 
 distinct.addAll(oldStringSet);
 distinct.addAll(newStringSet);

To enhance your code you may create a generic method for that

public static <T> Set<T> distinct(Collection<T>... lists) {
	Set<T> distinct = new HashSet<T>();
	
	for(Collection<T> list : lists) {
		distinct.addAll(list);
	}
	return distinct;
}

If you are using Guava you can also use a builder to get more flexibility:

ImmutableSet.<String>builder().addAll(someSet)
                              .addAll(anotherSet)
                              .add("A single string")
                              .build();

If you are using the Apache Common, use SetUtils class from org.apache.commons.collections4.SetUtils;

SetUtils.union(setA, setB);

Just use newStringSet.addAll(oldStringSet). No need to check for duplicates as the Set implementation does this already.

http://docs.oracle.com/javase/7/docs/api/java/util/Set.html#addAll(java.util.Collection)

Since sets can't have duplicates, just adding all the elements of one to the other generates the correct union of the two.

 newStringSet.addAll(oldStringSet);

This will produce Union of s1 and s2

Set.addAll()

Adds all of the elements in the specified collection to this set if they're not already present (optional operation). If the specified collection is also a set, the addAll operation effectively modifies this set so that its value is the union of the two sets

newStringSet.addAll(oldStringSet)

Use boolean addAll(Collection<? extends E> c)
Adds all of the elements in the specified collection to this set if they're not already present (optional operation). If the specified collection is also a set, the addAll operation effectively modifies this set so that its value is the union of the two sets. The behavior of this operation is undefined if the specified collection is modified while the operation is in progress.

newStringSet.addAll(oldStringSet)

If you care about performance, and if you don't need to keep your two sets and one of them can be huge, I would suggest to check which set is the largest and add the elements from the smallest.

Set<String> newStringSet = getNewStringSet();
Set<String> oldStringSet = getOldStringSet();

Set<String> myResult;
if(oldStringSet.size() > newStringSet.size()){
    oldStringSet.addAll(newStringSet);
    myResult = oldStringSet;
} else{
    newStringSet.addAll(oldStringSet);
    myResult = newStringSet;
}

In this way, if your new set has 10 elements and your old set has 100 000, you only do 10 operations instead of 100 000.