解法

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> countMap = new HashMap(nums.length);    
        for (int i : nums) {
            countMap.put(i, countMap.getOrDefault(i, 0) + 1);
        }
        // 小顶堆，第一个元素代表数字，第二个元素代表数字出现的次数
        PriorityQueue<int []> pq = new PriorityQueue<int []>(new Comparator<int []> () {
            public int compare(int[] n1, int[] n2) {
                return n1[1] - n2[1];
            }
        }); 

        for (Map.Entry<Integer, Integer> entry : countMap.entrySet()) {
            int val = entry.getKey();
            int count = entry.getValue();
            if (pq.size() == k) {
                // 等于k以后，如果堆顶的数目小于count,出堆
                if (pq.peek()[1] < count) {
                    pq.poll();
                    pq.offer(new int[]{val, count});
                }
            } else {
                pq.offer(new int[]{val, count});
            }
        }

        int[] ret = new int[k];
        int i = 0;
        while (pq.size() > 0) {
            ret[i] = pq.poll()[0];
            i++;
        }
        return ret;
    }
}