Bitwise ORs of Subarrays

题目
We have an array A of non-negative integers.

For every (contiguous) subarray B = [A[i], A[i+1], ..., A[j]] (with i <= j), we take the bitwise OR of all the elements in B, obtaining a result A[i] | A[i+1] | ... | A[j].

Return the number of possible results. (Results that occur more than once are only counted once in the final answer.)

答案

class Solution {
    public int subarrayBitwiseORs(int[] A) {
        Set<Integer> result_set = new HashSet<>();
        Set<Integer> lastline_set = new HashSet<>();
        
        for(int a : A) {
            Set<Integer> currline_set = new HashSet<>();
            currline_set.add(a);
            for(int s : lastline_set) {
                currline_set.add(s | a);
            }
            for(int s : currline_set) {
                result_set.add(s);
            }
            lastline_set = currline_set;
        }
        return result_set.size();
    }
}