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(shù)字A異或B兩次放坏，就得到A纽哥。而B(niǎo)被A異或兩次，就得到B
任意數(shù)與自己異或操作都是0，任意數(shù)與0異或操作都得到本身愚战。
#!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
