347. Top K Frequent Elements

Posted 2020-08-10 Machelsky

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<Integer> topKFrequent(int[] nums, int k) {
        List<Integer> save=new ArrayList<Integer>();
        if(nums.length==0)
        {
            return save;
        }
        Map<Integer,Integer> res=new HashMap<Integer,Integer>();
        for(int i=0;i<nums.length;i++)
        {
         if(!res.containsKey(nums[i]))
         {
             res.put(nums[i],1);
         }
         else
         {
             res.put(nums[i],res.get(nums[i])+1);
         }
        }
           
        PriorityQueue<Map.Entry<Integer, Integer>> maxHeap = 
                         new PriorityQueue<>((a,b)->(b.getValue()-a.getValue()));
        for(Map.Entry<Integer,Integer> entry:res.entrySet())
        {
            maxHeap.add(entry);
        }
        while(save.size()<k)
        {
            Map.Entry<Integer, Integer> entry = maxHeap.poll();
            save.add(entry.getKey());
        }
    
        return save;
    }
}