[leetcode]347. Top K Frequent Elements 最高频的前K个元素

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.

题意：

给定数组，求其中出现频率最高的K个元素。

Solution: PriorityQueue

扫数组，将数组每个元素和其对应的出现频率存到Map里

扫Map，将Map里的元素放入maxHeap

从maxHeap里poll出K个元素即可

code

 class Solution {
     public List<Integer> topKFrequent(int[] nums, int k) {
         Map<Integer, Integer> numFreqs = new HashMap();

         for(int num : nums){
             numFreqs.put(num, numFreqs.containsKey(num) ? numFreqs.get(num) + 1 : 1);
         }

         PriorityQueue <Map.Entry<Integer, Integer>> maxHeap = new PriorityQueue<>((entry1, entry2) -> entry2.getValue() - entry1.getValue());

         for(Map.Entry<Integer, Integer> entry : numFreqs.entrySet()){
             maxHeap.add(entry);
         }

         List<Integer> result = new ArrayList<>();
         while(!maxHeap.isEmpty() && result.size() < k){
             result.add(0, maxHeap.poll().getKey());
         }
         return result;

     }
 }