[leetcode-136-Single Number]

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