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I am trying to create a list, which only consists of unique values.

String arr = "5", "5", "7", "6", "7", "8", "0";
 List<String> uniqueList = new ArrayList<String>(new HashSet<String>( Arrays.asList(arr) ));
 System.out.println( uniqueList );

What I expect as an output is: 6,8,0. So, if duplicates exist, I want to delete both of them. The HashSet only removes the duplicates, so that each value only occurs once. However, I want to remove both the numbers, so that I end up with a list, which only has the numbers that occur once in the original list.

One solution would be to build a frequency Map and only retain the keys whose value equals 1:

String arr = "5", "5", "7", "6", "7", "8", "0";

Arrays.stream(arr)
 .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
 .entrySet()
 .stream()
 .filter(e -> e.getValue() == 1)
 .map(Map.Entry::getKey)
 .collect(Collectors.toList()));

One possible value of this List is:

[0, 6, 8]

Another possiblilty with Stream's:

List<String> arr1 = Arrays.asList(arr).stream()
 .filter(i -> Collections.frequency(Arrays.asList(arr), i) < 2)
 .collect(Collectors.toList());
arr1.forEach(System.out::println);

Which will create a filter out all the elements that occur more than once using Collections::frequency. Which returns the List:

[6, 8, 0]

One other possible solution is collecting the list data into set and then get back to list again.

String arr = "5", "5", "7", "6", "7", "8", "0";

List<String> stringList = Arrays.stream(arr).collect(Collectors.toSet()).stream().collect(Collectors.toList());

for (String s : stringList) 
 System.out.println(s);