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;
}