Would this be the correct method to check whether an integer array contains duplicates? I wanted to pass in an int[] nums instead of Integer[], but couldn't get that to work.
public static boolean isUnique(Integer[] nums){
return new HashSet<Integer>(Arrays.asList(nums)).size() == nums.length;
}
You can do something like:
public static boolean isUnique(int[] nums){
Set<Integer> set = new HashSet<>(nums.length);
for (int a : nums) {
if (!set.add(a))
return false;
}
return true;
}
This is more of a short-circuit-esque approach than what you have, returning as soon as it encounters a duplicate. Not to mention it works with an int[] as you wanted. We are exploiting the fact that Set#add returns a boolean indicating whether the element being added is already present in the set.
Whether Set or sorting is irrelevant here, and sorting is more optimal, less objects.
public static boolean isUnique(int[] nums) {
if (nums.length <= 1) {
return true;
}
int[] copy = Arrays.copyOf(nums);
Arrays.sort(copy);
int old = Integer.MAX_VALUE; // With at least 2 elems okay.
for (int num : copy) {
if (num == old) {
return false;
}
old = num;
}
return true;
}
Addendum As commented slower, though saving memory.
O(n*log(n)), whereas adding stuff to a HashSet is performed in O(n). For large arrays, sorting is certainly worse...
HashSet and its entries will escape eden space, so the GC can probably collect the whole HashSet shortly after leaving the method. The JVM has become pretty good at garbage collection... Note that the OP already has an Integer[] input array, not an int[] array