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?

異或交換原理: 數字A異或B兩次，就得到A。而B被A異或兩次，就得到B

任意數與自己異或操作都是0，任意數與0異或操作都得到本身。

#!usr/bin/env  
# -*-coding:utf-8 -*-
class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = 0
        for ele in nums:
            res ^= ele
        return res