Given a non-empty 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?

Example 1:

Input: [2,2,1]
Output: 1
Example 2:

Input: [4,1,2,1,2]
Output: 4

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/single-number
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这道题用异或做太火了。。。。

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int num=0;
        for(auto it :nums){
            num ^= it;
        }
        return num;
    }
};

想了一个，排序然后俩俩一对儿

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int n =nums.size();
        if(n==1){
            return nums[0];
        }
        sort(nums.begin(),nums.end());
        int i=0;
        while(i+1<n){
            if(nums[i]!=nums[i+1]){
                return nums[i];
            }
            i=i+2;
        }
        return nums[i];
    }
};