Java ( Counting Distinct Integers )

Question

例如，如何返回数组中不同/唯一值的数量

int[] a = {1,2,2,4,5,5};

Answer 1

Set<Integer> s = new HashSet<Integer>();
for (int i : a) s.add(i);
int distinctCount = s.size();

Answer 2

A set stores each unique (as defined by .equals()) element in it only once, and you can use this to simplify the problem. Create a Set (I'd use a HashSet), iterate your array, adding each integer to the Set, then return .size() of the Set.

Answer 3

An efficient method: Sort the array with Arrays.sort . Write a simple loop to count up adjacent equal values.

Answer 4

Really depends on the numbers of elements in the array. If you're not dealing with a large amount of integers, a HashSet or a binary tree would probably be the best approach. On the other hand, if you have a large array of diverse integers (say, more than a billion) it might make sense to allocate a 2^32 / 2^8 = 512 MByte byte array in which each bit represents the existence or non-existence of an integer and then count the number of set bits in the end.

A binary tree approach would take n * log n time, while an array approach would take n time. Also, a binary tree requires two pointers per node, so your memory usage would be a lot higher as well. Similar consideration apply to hash tables as well.

Of course, if your set is small, then just use the inbuilt HashSet.