1310. XOR Queries of a Subarray

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

class Solution {
    public int[] xorQueries(int[] arr, int[][] queries) {
        int[] res = new int[queries.length];
        int ql = queries.length;
        for(int i = 0; i < ql; i++){
            res[i] = help(arr, queries[i][0], queries[i][1]);
        }
        return res;
    }
    public int help(int[] arr, int l, int r){
        int re = 0;
        if(l == r) return arr[l];
        while(l <= r){
            re = arr[l] ^ (arr[r] ^ re);
            l++;
            r--;
            if(l == r) return re ^ arr[l];
        }
        return re;
    }
}

0和任何数异或数值不变

然后就直接解，大概是O（n/2），注意判断l和r的大小关系