LeetCode 347. Top K Frequent Elements 前 K 个高频元素 (Java)

题目：

Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

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 nis the array's size.
• It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
• You can return the answer in any order.

分析：

找出前K个高频元素，TopK的问题一般都可以利用堆，或者优先级队列来处理，统计好每个元素出现的顺序，加入到容量大小为k的堆中，即可得到答案。

程序：

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int num:nums){
            int times = map.getOrDefault(num, 0);
            map.put(num, times+1);
        }
        PriorityQueue<Integer> minHeap = new PriorityQueue<>((a, b)->{
            return map.get(a) - map.get(b);
        });
        for(int num:map.keySet()){
            minHeap.offer(num);
            if(minHeap.size() > k)
                minHeap.poll();
        }
        int[] res = new int[k];
        for(int i = 0; i < res.length; ++i){
            res[i] = minHeap.poll();
        }
        return res;
    }
}