single-number

Given an array of integers, every element appears twice except for one. Find that single one.

Note: 
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

import java.util.*;
import java.util.Map.*;
public class Solution {
    public int singleNumber(int[] A) {
      /*  HashMap<Integer,Integer> map=new HashMap<>();
        for(int i=0;i<A.length;i++){
            if(map.get(A[i])==null)
                map.put(A[i],1);
            else map.put(A[i],map.get(A[i])+1);
        }
        
        for(Entry<Integer,Integer> entry:map.entrySet()){
            int a=entry.getValue();
            int k=entry.getKey();
            if(a==1)
                return k;
        }
        
        return 0;
        */
        
        //使用异或，两个二进制数异或，相同就会为0,不同的为1。所以两个相同整数异或为0，而所有整数与0异或还是这个整数本身
        int result=A[0];
        for(int i=1;i<A.length;i++)
            result^=A[i];
        return result;
        
    }
}