So I want to insert random integers to a list. However, I want to make sure that there are no duplicates on the list. Here is what I have done so far:
while (options.size() <=4) {
int i = 0;
int random = ThreadLocalRandom.current().nextInt(answer - 8 ,answer + 8);
if (random != answer && random != options.get(i-1)) {
options.add(random);
i++;
}
}
The easiest way is with a distinct random stream:
ThreadLocalRandom.current()
.ints(answer - 8, answer + 8)
.distinct()
.limit(5)
.forEach(options::add);
For better performance on bigger datasets, you can shuffle instead:
List<Integer> range = IntStream.range(answer - 8, answer + 8)
.boxed()
.collect(Collectors.toCollection(ArrayList::new));
Collections.shuffle(range);
options.addAll(range.subList(0, 5));
Since you want distinct values in your collection, use a LinkedHashSet and add values in it.
Set<Integer> options = new LinkedHashSet<>();
while (options.size() <= 4) {
int i = 0;
int random = ThreadLocalRandom.current().nextInt(answer-8, answer+8);
if (random != answer) {
options.add(random);
i++;
}
}
Once done, you can convert the set to a list (if required for further processing)
List<Integer> list = set.stream().collect(Collectors.toList());
It turns out all I needed was the contain function
while (options.size() <=4){
int i = 0;
int random = ThreadLocalRandom.current().nextInt(answer -8 ,answer +8 );
if (random != answer && !options.contains(random)){
options.add(random);
i++;
}
}
contains on a List is O(n) - that makes this algorithm a hugely inefficient O(n^2). Use a LinkedHashSet instead.