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?

学习到了异或的一个特性：

同一变量与另一变量和其异或值异或等于另一个数，如（a^b）^a=b。

这样，将所有元素依次与第一个值做异或操作，便能够得到唯一的那个值。

int singleNumber(vector
     
      & nums) {      for (int i = 1; i < nums.size();i++)      {           nums[0] ^= nums[i];       }      return nums[0]; }