1041. Be Unique (20)

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1, 10⁴]. The first one who bets on a unique number wins. For example, if there are 7 people betting on 5 31 5 88 67 88 17, then the second one who bets on 31 wins.

Input Specification:

Each input file contains one test case. Each case contains a line which begins with a positive integer N (<=10⁵) and then followed by N bets. The numbers are separated by a space.

Output Specification:

For each test case, print the winning number in a line. If there is no winner, print "None" instead.

Sample Input 1:

7 5 31 5 88 67 88 17

Sample Output 1:

31

Sample Input 2:

5 888 666 666 888 888

Sample Output 2:

None

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

int main()
{
	int N, num;
	vector<int>vec;
	int res = 0;
	scanf("%d",&N);
	for (int i = 0; i < N; ++i)
	{
		scanf("%d",&num);
		vec.push_back(num);
	}

	for (int i = 0; i < vec.size(); ++i)
	{
		int j = 0;
		for (j = 0; j < vec.size(); ++j)
		{
			if (i == j)
				continue;
			if (vec[i] == vec[j])
			{
				break;
			}
		}
		if (j == vec.size())
		{
			res = vec[i];
			break;
		}
	}
	if (res!=0)
		printf("%d", res);
	else
		printf("%s","None");
}

上面的代码还有两个小测试点是超时的，也就是还有优化的空间. 通过加入一个辅助的set,记录下检测过的已经重复的元素，然后再检测当前元素之前进行一个判断，看当前元素是属于辅助的set，也就是说不进行相同元素的重新检测，因此问题得到的优化，测试全部通过.

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

int main()
{
	int N, num;
	vector<int>vec;
	set<int> s;
	int res = 0;
	scanf("%d", &N);
	for (int i = 0; i < N; ++i)
	{
		scanf("%d", &num);
		vec.push_back(num);
	}

	for (int i = 0; i < vec.size(); ++i)
	{
		set<int>::iterator s_it = s.find(vec[i]);
		if (s_it != s.end())
			continue;
		int j = 0;
		for (j = i+1; j < vec.size(); ++j)
		{
			if (vec[i] == vec[j])
			{
				s.insert(vec[i]);
				break;
			}
		}
		if (j == vec.size())
		{
			res = vec[i];
			break;
		}
	}
	if (res != 0)
		printf("%d", res);
	else
		printf("%s", "None");
}