【leetcode】Single Number (Medium) ☆

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?

 

要求O(n)算法，还不能有辅助空间。开始用了各种O(n^2)的方法，都超时，后来突然顿悟了。异或可以消掉两个相同数字。

直接把所有数字异或在一起，就是单独的那个数字了。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public://异或可以把两个相同的数字消除 O（n) AC
    int singleNumber3(int A[], int n) {
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            ans ^= A[i];
        }
        return ans;
    }
};

int main()
{
    Solution s;
    int a[9] = {0,0,1,1,2,3,2,3,4};
    int n = s.singleNumber3(a, 9);

    return 0;
}