Leecode subarray XOR query

There's an array of positive integers arr, Now I give you a corresponding query array queries, among queries[i] = [Li, Ri].

For each query i, Please calculate from Li To Ri Of XOR value （ namely arr[Li] xor arr[Li+1] xor … xor arr[Ri]） As a result of this query .

And returns a query containing the given queries An array of all the results .

Example 1：

 Input ：arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
 Output ：[2,7,14,8] 
 explain ：
 The binary representation of the elements in an array is ：
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
 Of the query  XOR  The value is ：
[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]

Tips ：
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

Ideas ：

	1、queries The first and second elements of each row in the array of , Namely arr The left and right bounds of the array 
	2、 stay queries Take the first and second numbers of a row of data in the array as the left and right bounds , Starting from the left boundary, it is connected with the next number in turn, or , Give the result to ans In the array 
	3、 return ans

Implementation code ：

/* The following code is my local Idea For testing in environment  public class XorSelect { public static void main(String[] args) { int arr[]={4,8,2,10}; int [][]q={
    {2,3},{1,3},{0,0},{0,3}}; int[] ints = new Solution().xorQueries(arr, q); } } */

class Solution{
    
    public int[] xorQueries(int[] arr, int[][] queries) {
    
        int []ans=new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
    
                int l=queries[i][0];
                int r=queries[i][1];


                    int res=arr[l];
                for (int j=l;j<r;j++){
    
                    res=res^arr[j+1];
                }

        /* if (i!= queries.length-1)System.out.print(res+" " ); else System.out.println(res); */
            ans[i]=res;
        }

        return ans ;
    }
}