题目描述

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.)

Example 1:

Input: [0]
Output: 1
Explanation: 
There is only one possible result: 0.

Example 2:

Input: [1,1,2]
Output: 3
Explanation: 
The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
These yield the results 1, 1, 2, 1, 3, 3.
There are 3 unique values, so the answer is 3.

Example 3:

Input: [1,2,4]
Output: 6
Explanation: 
The possible results are 1, 2, 3, 4, 6, and 7.

Note:

1 <= A.length <= 50000 0 <= A[i] <= 10^9

思路

每增加一个数字，就是在之前的所有序列中，新增一个数字，以及该数字和之前所有序列的 或| 值。 其实也不难想到，为啥我就想不到呢。。。。 三天没做题，又开始没思路了。。。嘤。

代码

class Solution {
          
   
public:
    int subarrayBitwiseORs(vector<int>& A) {
          
   
        unordered_set<int> cur, res, tmp;
        for (const auto& num1 : A) {
          
   
            tmp.clear();
            tmp = {
          
   num1};
            for (const auto& num2 : cur) {
          
   
                tmp.insert(num1 | num2);
            }
            cur = tmp;
            res.insert(cur.begin(), cur.end());
        }
        return res.size();
    }
};