Find out only 1 occurrences of the number in the array

Any number XOR itself equals 0, two repetitions of the number XOR, and then only 1 digits left.
Knowledge Point: Any number XOR 0 equals itself
Bitwise AND A&B only the two sides are 1 for 1
Bitwise OR | One of the 1 is 1.
Bitwise XOR ^ The same for both sides is 0 different for 1
Note: This method is only suitable for other numbers to appear even if the number of times

public class Test_plus {    public static int findNotDouble(int[] a) {        int n=a.length;        int result=a[0];        int i;        for(i=1;i<n;++i)            result ^= a[i];        return result;    }    public static void main(String[] args) {        int array[]= {1,2,3,2,4,3,5,4,1};        int num = findNotDouble(array);        System.out.println(num);    }}

Find out only 1 occurrences of the number in the array