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?

一刷
題解：
異或滿足交換律，且A ^ A = 0
A ^ B ^ C = B ^ A ^ C
那么一路異或下來，可以找到single number

public class Solution {
    public int singleNumber(int[] nums) {
        int res = 0;
        for(int num : nums){
            res ^= num;
        }
        return res;
    }
}