1310 XOR Queries of a Subarray 子数组异或查询
Description:
You are given an array arr of positive integers. You are also given the array queries where queries[i] = [lefti, righti].
For each query i compute the XOR of elements from lefti to righti (that is, arr[lefti] XOR arr[lefti + 1] XOR ... XOR arr[righti] ).
Return an array answer where answer[i] is the answer to the ith query.
Example:
Example 1:
Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
Output: [2,7,14,8]
Explanation:
The binary representation of the elements in the array are:
1 = 0001
3 = 0011
4 = 0100
8 = 1000
The XOR values for queries are
[0,1] = 1 xor 3 = 2
[1,2] = 3 xor 4 = 7
[0,3] = 1 xor 3 xor 4 xor 8 = 14
[3,3] = 8
Example 2:
Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
Output: [8,0,4,4]
Constraints:
1 <= arr.length, queries.length <= 3 * 10^4
1 <= arr[i] <= 10^9
queries[i].length == 2
0 <= lefti <= righti < arr.length
题目描述：
有一个正整数数组 arr，现给你一个对应的查询数组 queries，其中 queries[i] = [Li, Ri]。
对于每个查询 i，请你计算从 Li 到 Ri 的 XOR 值（即 arr[Li] xor arr[Li+1] xor ... xor arr[Ri]）作为本次查询的结果。
并返回一个包含给定查询 queries 所有结果的数组。
示例：
示例 1：
输入：arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
输出：[2,7,14,8]
解释：
数组中元素的二进制表示形式是：
1 = 0001
3 = 0011
4 = 0100
8 = 1000
查询的 XOR 值为
[0,1] = 1 xor 3 = 2
[1,2] = 3 xor 4 = 7
[0,3] = 1 xor 3 xor 4 xor 8 = 14
[3,3] = 8
示例 2：
输入：arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
输出：[8,0,4,4]
提示：
1 <= arr.length <= 3 * 10^4
1 <= arr[i] <= 10^9
1 <= queries.length <= 3 * 10^4
queries[i].length == 2
0 <= queries[i][0] <= queries[i][1] < arr.length
思路：
前缀和
先求出异或运算的前缀和 xors[i] = arr[:i] 的异或和
查询的时候可以使用差分运算直接求出区间的异或和
时间复杂度为 O(n), 空间复杂度为 O(n)
代码:
C++:
class Solution
{
public:
vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries)
{
int n = arr.size(), m = queries.size();
vector<int> xors(n + 1), result(m);
for (int i = 0; i < n; i++) xors[i + 1] = xors[i] ^ arr[i];
for (int i = 0; i < m; i++) result[i] = xors[queries[i].front()] ^ xors[queries[i].back() + 1];
return result;
}
};
Java:
class Solution {
public int[] xorQueries(int[] arr, int[][] queries) {
int[] xors = new int[arr.length + 1], result = new int[queries.length];
for (int i = 0; i < arr.length; i++) xors[i + 1] = xors[i] ^ arr[i];
for (int i = 0; i < queries.length; i++) result[i] = xors[queries[i][0]] ^ xors[queries[i][1] + 1];
return result;
}
}
Python:
class Solution:
def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
xors = [0]
for num in arr:
xors.append(xors[-1] ^ num)
return [xors[left] ^ xors[right + 1] for left, right in queries]
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