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.

 

思路：其实思路很简单。扫一遍array，把数和次数存到hashmap。然后按照次数大的priority把entry存到prioirty queue里面。最后按照k的个数依次从priority queue里面读出来。有Map.Entry这个神奇包裹省事很多。

Map.Entry:https://docs.oracle.com/javase/7/docs/api/java/util/Map.Entry.html

public class Solution {    public List
     
       topKFrequent(int[] nums, int k) {        List
      
        save=new ArrayList
       
        ();        if(nums.length==0)        {            return save;        }        Map
        
          res=new HashMap
         
          ();        for(int i=0;i
          
           > maxHeap = new PriorityQueue<>((a,b)->(b.getValue()-a.getValue())); for(Map.Entry
           
             entry:res.entrySet()) { maxHeap.add(entry); } while(save.size()
            
              entry = maxHeap.poll(); save.add(entry.getKey()); } return save; }}