[C ++] - the number of fruit with a map appears statistics

The document questions before CVTE there have been such a problem, generally means that the statistics of the number of fruits appear, and then come before the k kinds of employees like to eat the fruit.
That problem here can be divided into two parts

- a first count is the number of times each fruit appears

There are three ways:

//第一种：
void test_map1()
{
	string strs[]={"苹果"，"香蕉"，"香蕉","苹果"，"香蕉"，"苹果"，"香蕉"，"香蕉"，"草莓"};
	map<string,int> countmap;
	for(auto str:strs)
	{
		auto ret=countmap.find(str);
		if(ret!=countmap.end())
		{
			ret->second++;
		}
		else
		{
			countmap.insert(pair<string,int>(str,1));
		}
	}
}
//第二种方法
void test_map2()
{
	string strs[] = { "苹果","香蕉","香蕉","苹果","香蕉","苹果","香蕉","香蕉","草莓" };
	map<string, int> countmap;
	for (auto str : strs)
	{
		//pair<map<string, int>::iterator, bool> ret = countmap.insert(pair<string, int>(str, 1));
		auto ret= countmap.insert(pair<string, int>(str, 1));  //无论是什么水果都进行插入
		if (ret.second == false)
		{
			ret.first->second++;
		}
	}
}
//第三种方法
void test_map3()
{
	string strs[] = { "苹果","香蕉","香蕉","苹果","香蕉","苹果","香蕉","香蕉","草莓" };
	map<string, int> countmap;
	for (auto str : strs)
	{
		countmap[str]++;
	}
}

The first two methods will be well understood, for the third method is mainly overloaded "[]", it would look a bit complicated, function prototype is as follows:

mapped_type& operator[] (const key_type&k)
{
	return (*((this->insert(make_pair(k, mapped_type()))).first)).second;
}

May be difficult to understand, then I drew a diagram, and finally returns iterator iterator, then dereference it returned iterator is an alias, and then take it the second it is easy to get value for value.

- The second step is to issue a topk
number of occurrences of fruit are sorted in descending order, remove the first k elements

struct compare
{
	bool operater()(const pair<string, int>&l, const pair < string, int)&r)
	{
	return l.second > r.second;
	}
};
void test_map3()
{
	string strs[] = { "苹果","香蕉","香蕉","苹果","香蕉","苹果","香蕉","香蕉","草莓" };
	map<string, int> countmap;
	for (auto str : strs)
	{
		countmap[str]++;
	}
	vector<pair<string, int>> v;
	for (auto &e : countmap)
	{
		v.push_back(e);
	}
	sort(v.begin(), b.end(), compare());
	//将结果打印出来
	for (size_t i = 0; i < k; i++)
	{
		cout << v[i].first << ":" << v[i].second << endl;
	}
}

• Here are a priority queue solution again

//利用优先级队列解决
void GetFavoriteFruit(const vector<string>&fruits, size_t k)
{
	//首先统计每种水果的次数
	map<string, int> countmap;
	for (auto &e : fruits)
	{
		countmap[e]++;
	}
   // 排序，先让前k种水果进入队列，然后大于k的在再次比较他们的次数
	priority_queue < pair<string, int>, vector<pair<string, int>, compare>> pq;
	size_t i = 0;
	for (auto e : countmap)
	{
		if (i < k)
		{
			pq.push(e);
			++i;
		}
		else
		{
			if (e.second > pq.top().second)
			{
				pq.pop();
				pq.push(e);
			}
		}
	}
	
}