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.)

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

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.

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

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

1. 动态规划：遍历数组，记录以每一个数为结尾的子数组的结果，存在set中；
2. 将当前数与前一个数的结果的取或和本身，存储aSet（set）中；
3. aSet的长度即使所有结果的可能数。

class Solution:
def subarrayBitwiseORs(self, A):
    index =A[0]
    aSet=set()
    judge=set()
    judge.add(index)
    aSet.add(index)
    for i in range(1,len(A)):
        _judge=set()
        for j in judge:
            aSet.add(A[i])
            _judge.add(A[i])
            aSet.add(j | A[i])
            _judge.add(j | A[i])
        judge=_judge
    return len(aSet)

83 / 83 test cases passed.
Runtime: 1712 ms