I have a set
like below:
HashSet<Integer> set = new HashSet<Integer>();
set.add(1);
How can I get the 1
out? I can do it by for(integer i : set)
. My specified problem is "Given an array of integers, every element appears twice except for one. Find that single one."
I want to use add elements into the set if the set doesn't contain it and remove existing elements during the loop. And the last remaining element is the answer. I don't know how to return it.
public static int singleNumber(int[] A) {
HashSet<Integer> set = new HashSet<>();
for (int a : A) {
if (!set.contains(a)) {
set.add(a);
} else {
set.remove(a);
}
}
/**
* for(Integer i : set) { return i; }
*return A[0]; //need one useless return
/**
* while(set.iterator().hasNext()) { return set.iterator().next(); }
* return A[0]; //need one useless return
*/
return set.toArray(new Integer[1])[0];
}
Simply try using HashSet#toArray()
method
HashSet<Integer> set = new HashSet<Integer>();
set.add(1);
if (set.size() == 1) { // make sure that there is only one element in set
Integer value = set.toArray(new Integer[1])[0];
System.out.println(value);//output 1
}