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?在数组中找到唯一一个没有两次出现的整数，要求线性的时间(O(n)),没有额外的空间消耗不能的hash表，hash需要额外的空间；采用异或操作a^a=0a^b^a=b;数组中只有一个数出现一次，其他的数都出现两次，异或后的结果就是那个单次出现的整数*/class Solution {public:    int singleNumber(vector<int>& nums) {        int res=nums[0];        for(int i=1;i<nums.size();i++)            res^=nums[i];        return res;    }};