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?

解題思路：使用異或，數(shù)組中相同的數(shù)兩兩異或全部為0，最后剩下的就是那個(gè)落單的數(shù)。

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