LeetCode136：Single Number

题目：

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?

解题思路：

很简单的一题，直接用异或运算解决：连个相同数进行异或结果为0,0与任何数异或结果为原数

也可先进行排序再遍历排序后的数组，当某个数没有重复时，结果就是它了

也可用bitmap进行，不多说了。

实现代码：

#include <iostream>

using namespace std;
/*
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?
*/
class Solution {
public:
    int singleNumber(int A[], int n) {
        int ret = 0;
        for(int i = 0; i < n; i++)
            ret ^= A[i];
        return ret;
        
    }
};

int main(void)
{
    int arr[] = {2,4,5,5,4,1,2};
    int len = sizeof(arr) / sizeof(arr[0]);
    Solution solution;
    int once = solution.singleNumber(arr, len);
    cout<<once<<endl;
    return 0;
}