SingleNumber

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?

//意思是一堆数中找出没成对的那个，用异或效率最高A^0=A   A^A=0 最后可求出SingleNumber

public class Solution {

    public int singleNumber(int[] A) {
        int res=0;
        for(int i=0;i<A.length;i++){
            res^=A[i];
        }
        return res;
    }
}