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?

解题思路：

这道题主要考察位运算，x ^ x = 0，数组中只有一个数字出现一次，其他数字都出现了两次，对所有数字进行异或运算后，将得到只出现一次的那个数。

代码：

#include <iostream>#include <vector>using namespace std;int singleNumber(vector<int>& nums){     int sum = 0, len = nums.size();     for(int i = 0; i < len; i ++)     {         sum = sum ^ nums[i];     }     return sum;}int main(){    vector<int> nums;    int num;    while(cin>>num)    {        nums.push_back(num);    }    cout<<singleNumber(nums);}