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

分析：

找出陣列中只出現一次的數字，用異或2次會互相抵消的特性，當將數組裡面的所有數異或一遍後，結果只會留下那個只出現1次的。

程式碼：

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int n=nums[0];
        for(int i=1;i<nums.size();++i)
            n=n^nums[i];
        return n;
    }
};