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?

思路：

这里使用异或运算，异或运算有如下性质：

1.相同元素异或为0,0与任何数异或等于任何数，有a^b^a=b。

2.此外，异或还可以用于两个元素交换a=a^b^(b=a)。

实现：

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def singleNumber(nums):
        res = 0
        for i in nums:
            res = res^i
        return res