#include <iostream>#include <stdio.h>#include <vector>#include <string>using namespace std;/*問題: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?分析:一個數(shù)組中，每個元素都出現(xiàn)了兩次，只有一個元素不是。這明顯是用異或來做的，根據(jù)b=a^a^b輸入:31 1 2輸出:2*/class Solution {public:    int singleNumber(vector<int>& nums) {        if(nums.empty())		{			return 0;		}		int size = nums.size();		int result = 0;		for(int i = 0 ; i < size ; i++)		{			result ^= nums.at(i);		}		return result;    }};void PRint(vector<int>& result){	if(result.empty())	{		cout << "no result" << endl;		return;	}	int size = result.size();	for(int i = 0 ; i < size ; i++)	{		cout << result.at(i) << " " ;	}	cout << endl;}void process(){	 vector<int> nums;	 int value;	 int num;	 Solution solution;	 vector<int> result;	 while(cin >> num )	 {		 nums.clear();		 for(int i = 0 ; i < num ; i++)		 {			 cin >> value;			 nums.push_back(value);		 }		 num = solution.singleNumber(nums);		 cout << num << endl;	 }}int main(int argc , char* argv[]){	process();	getchar();	return 0;}