1310. XOR Queries of a Subarray ------M

 Given the array arr of positive integers and the array queries where queries[i] = [Li, Ri], for each query i compute the XOR of elements from Li to Ri (that is, arr[Li] xor arr[Li+1] xor ... xor arr[Ri] ). Return an array containing the result for the given queries.

 

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 <= 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

A:

类似于 subarray sum.  每个地方保存之前的值。

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class Solution { public: vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) { int n = arr.size(); vector<int> V(n+1,0); for(int i = 0;i<n;i++){ V[i+1] = V[i] ^ arr[i]; } vector<int> res; for(int i =0;i<queries.size();i++){ int l = queries[i][0]; int r = queries[i][1]; res.push_back(V[r+1] ^ V[l]); } return res; } };