思路

这道题还是使用优先队列，是要大根堆，然后创建一个类，成员变量值和次数。大根堆基于次数排序。前k个就拿出前k的类的值即可。代码如下：

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k < 1 || k > nums.length) {
            return null;
        }
        int[] ans = new int[k];
        PriorityQueue<Info> priorityQueue = new PriorityQueue<>((o1, o2) -> o2.times - o1.times);
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            if (map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            } else {
                map.put(num, 1);
            }
        }
        map.forEach((value, times) -> {
            Info info = new Info();
            info.times = times;
            info.value = value;
            priorityQueue.add(info);
        });

        for (int i = 0; i < k; i++) {
           ans[i] = priorityQueue.poll().value;
        }
        return ans;
    }

    class Info {
        public int times;
        public int value;

        public Info() {
        }
    }
}