136. Single Number

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?
Analysis:
这道题可使用异或运算求解，主要用到以下性质：
1.异或运算符合交换律和结合律。即a^b=b^a; (a^b)^c=(a^b)^c
2.a^a=0
补充：异或运算的其他使用：使特定位翻转（^1）,使特定位保持不变（^0）
Source Code（C++）:

#include <iostream>#include <vector>using namespace std;class Solution {public:    int singleNumber(vector<int>& nums) {               if (nums.empty()){            return -1;        }        int single_number = nums.at(0);        for (int i=1; i<nums.size(); i++){            single_number ^= nums.at(i);        }        return single_number;    }};int main() {    Solution sol;    vector<int> v;    cout << sol.singleNumber(v);    return 0;}