leetcode898.BitwiseORsofSubarrays

参考：https://blog.csdn.net/u014110320/article/details/82316690
参考：https://blog.csdn.net/yeshen4328/article/details/82563040

题目链接

898. Bitwise ORs of Subarrays

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.)

Example1

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

Example2

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.

Example3

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

暴力方法

class Solution {public int subarrayBitwiseORs(int[] A) {if(A == null || A.length == 0) {return 0;}Set<Integer> set = new HashSet<>();int max = 0;for(int i = 0; i < A.length; i++) {set.add(A[i]);max |= A[i];set.add(max);}for(int i = 0; i < A.length; i++) {for(int j = i+1, n = 0; j < A.length; j++) {n |= A[j];set.add(n);if(n == max) {break;}}} return set.size();}
}

我们可以首先找出以Ai结尾的所有字串，每一个子串都可以计算一个结果（可能有些是重复的），因为Ai是4字节整数，有32位，而且这一行结果中，左边的结果比右边结果的二进制多一个“1”或者相等（因为左边结果比右边结果多“或”了一个或者多个元素），如果每一个结果都是不一样的，那么这一行结果最大可能有32种：

从A1开始，初始化一个临时set，将Ai加入到set中，并用Ai和上一个临时set的所有元素做或运算，把结果存进全局set以及当前临时set中，然后进入下一个迭代。示意图如下：

class Solution {public int subarrayBitwiseORs(int[] A) {Set<Integer> set = new HashSet<Integer>() ;Set<Integer> last = new HashSet<Integer>() ;int len = A.length ;for(int i=0;i<len;i++){Set<Integer> temp = new HashSet<Integer>() ;temp.add(A[i]) ;for(int element : last){temp.add(A[i] | element) ;}last = temp ;for(int element :temp){set.add(element) ;}}return set.size() ;}
}