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;

    }

}