LC 898. Bitwise ORs of Subarrays

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.)

Runtime: 652 ms

Memory Usage: 49.7 MB

class Solution {
public:
  int subarrayBitwiseORs(vector<int>& A) {
    unordered_set<int> s;
    set<int> t;
    for(int i : A) {
      set<int> r;
      r.insert(i);
      for(int j : t) r.insert(j | i);
      t = r;
      for(int j : t) s.insert(j);
    }
    return s.size();
  }
};

原文地址：https://www.cnblogs.com/ethanhong/p/10352715.html