连接：https://leetcode-cn.com/problems/top-k-frequent-elements/solution/leetcode-di-347-hao-wen-ti-qian-k-ge-gao-pin-yuan-/

优先队列（堆排序）或者桶排序
注意一下对于ArrayList数组的创建：ArrayList<Integer>[] list = new ArrayList[nums.length + 1];

/**
 * topK问题，使用map记录每个元素出现的频率，然后使用小顶堆或者桶排序即可
 * https://leetcode-cn.com/problems/top-k-frequent-elements/solution/leetcode-di-347-hao-wen-ti-qian-k-ge-gao-pin-yuan-/
 */
public class topK {
    public static List<Integer> topKFrequent(int[] nums, int k) {
        List<Integer> res = new ArrayList<>();
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            if (!map.containsKey(num)) map.put(num, 1);
            else {
                map.put(num, map.get(num) + 1);
            }
        }
        PriorityQueue<Integer> queue = new PriorityQueue<>(new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return map.get(o1) - map.get(o2);
            }
        });
        for (Integer index : map.keySet()) {
            if (queue.size() >= k) {
                if (map.get(queue.peek()) < map.get(index)) {
                    queue.poll();
                    queue.offer(index);
                }
            } else {
                queue.offer(index);
            }
        }
        while (!queue.isEmpty()) {
            res.add(queue.poll());
        }
        return res;
    }


    public static List<Integer> topKFrequent2(int[] nums, int k) {
        List<Integer> res = new ArrayList<>();
        Map<Integer, Integer> map = new HashMap<>();
        ArrayList<Integer>[] list = new ArrayList[nums.length + 1];
        // 统计频率
        for (int num : nums) {
            if (!map.containsKey(num)) map.put(num, 1);
            else map.put(num, map.get(num) + 1);
        }
        // 向频率桶中加入相应的元素
        for (Integer index : map.keySet()) {
            int freq = map.get(index);
            if (list[freq] == null) list[freq] = new ArrayList<>();
            list[freq].add(index);
        }
        // 从频率最高的桶开始往前查找，找到k个数就结束
        for (int i = list.length - 1; i >= 0 && res.size() < k; i--) {
            if (list[i] == null) continue;
            while (list[i].size() != 0) {
                res.add(list[i].remove(0));
            }
        }
        return res;
    }

    public static void main(String[] args) {
        int[] nums = {1, 1, 1, 2, 2, 3};
        System.out.println(topKFrequent2(nums, 2));
    }
}