Leetcode 347. top k high frequency elements

Leetcode 347. front K High frequency elements

subject

Give you an array of integers nums And an integer k , Please return to the frequency before k High element . You can press In any order Return to the answer .

Ideas

• First, use the hash table to count the number of occurrences of each element
• Define a small root heap , It's actually a priority queue , The length is k, Send all frequencies into the stack , Build a small root heap , If the length is greater than k, Straight out of the team , Every time out of the team is the smallest element

Code

class Solution {
    
public:
    class mycomparison{
    
        public:
            bool operator()(const pair<int,int>& lhs,const pair<int,int>& rhs){
    
                return lhs.second > rhs.second;
            }
    };

    vector<int> topKFrequent(vector<int>& nums, int k) {
    
        //  First count the occurrence frequency of elements 
        unordered_map<int,int> map;//  Use a hash table to store the number of occurrences of each element 
        for(int i = 0; i < nums.size(); i++)
        {
    
            map[nums[i]]++;//  Add one to the number of times the element appears 
        }
        //  Sort frequencies 
        //  Define a small root heap   The size is k
        priority_queue<pair<int,int>,vector<pair<int,int>>,mycomparison> pri_que; 

        //  Use a fixed size of k Small roots 
        for(unordered_map<int,int>::iterator it = map.begin();it != map.end(); it++)
        {
    
            pri_que.push(*it);
            if(pri_que.size() > k)
            {
    
                //  If the heap size is greater than k  Queue Popup   Make sure the heap size is always K  Maintain a priority queue 
                pri_que.pop();
            }
        }

        //  Before finding out k Elements , The smallest element pops up first in the small root heap , So output in reverse order 
        vector<int> result(k);
        for(int i = k - 1; i >= 0; i--)
        {
    
            result[i] = pri_que.top().first;//  Get the first element of the queue 
            pri_que.pop();
        }

        return result;
    }
};