【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;

}