[leetcode] 898. Bitwise ORs of Subarrays
Description
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
分析
題目的意思是:給定一個陣列,求子陣列的異或運算得到的值的數目。這裡用三個set集合來計算,第一個set集合s用來儲存最終的值,prev儲存上一次遍歷計算得到的異或值,t儲存當前計算得到的異或值。然後遍歷維護這三個集合就行了。
程式碼
class Solution:
def subarrayBitwiseORs(self, A: List[int]) -> int:
s=set()
prev=set()
for x in A:
t=set()
for y in prev:
t.add(x|y)
t.add(x)
prev=t
s|=t
return len(s)
參考文獻
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