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：a^b^b=a以及异或操作的交换律。
代码：
1 class Solution { 2 public: 3 int singleNumber(vector<int>& nums) { 4 int Xor=0; 5 for(int i=0;i<nums.size();i++){ 6 Xor=Xor^nums[i]; 7 } 8 return Xor; 9 } 10 };
原文：http://www.cnblogs.com/Deribs4/p/5722445.html