347. Top K Frequent Elements

/*Given a non-empty array of integers, return the k most frequent elements.For example,Given [1,1,1,2,2,3] and k = 2, return [1,2].Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements.Your algorithm's time complexity must be better than O(n log n), where n is the array's size.使用一个hash表对重复的数字计数，使用优先队列对hash中的计数值进行排序*/#include <iostream>#include <vector>#include <unordered_map>#include <queue>#include <utility>using namespace std;class Solution {private:    struct bigger{        bool operator()(pair<int,int> &a,pair<int,int>&b)        {            return a.first > b.first;        }    };public:    vector<int> topKFrequent(vector<int>& nums, int k) {        unordered_map<int,int> mp;        for(int i=0;i<nums.size();i++)            mp[nums[i]]++;        vector<int> res;        priority_queue<pair<int,int>,vector<pair<int,int>>,bigger> p;//小顶堆        for(unordered_map<int,int>::iterator it=mp.begin();it!=mp.end();it++)        {            if(p.size()<k)                p.push(make_pair(it->second,it->first));            else if(it->second >= p.top().first)            {                p.pop();                p.push(make_pair(it->second,it->first));            }        }        while(!p.empty())        {            res.push_back(p.top().second);            p.pop();        }        return res;    }};int main(){    Solution mys;    vector<int> nums={2,3,4,3,2,3};    int k=2;    vector<int>res=mys.topKFrequent(nums,k);    for(int i=0;i<res.size();i++)        cout<<res[i]<<endl;    return 0;}