2.O(n), O(n). bucket sort。要用到频率最多也就是n的信息+O(n)的空间。

第一遍扫计算频率

第二遍扫根据频率把元素值放到对应frequency的bucket里。

第三遍扫bucket，从右往左（从高频到低频），凑满k个元素。

细节：

1.bucket因为不能保证某一频率只有一元素，所以要用List<Integer>[], generic type初始化特别，要List<Integer>[] buckets = new List[nums.length + 1];注意等号右边不带generic 

class Solution {private class EntryComparator implements Comparator<Map.Entry<Integer, Integer>> {@Override// P2: 要过OJ要声明一下Map.Entry, 没办法只写Entry，因为不够普遍吧，没适配。public int compare(Map.Entry<Integer, Integer> a, Map.Entry<Integer, Integer> b) {return a.getValue() - b.getValue();}}public List<Integer> topKFrequent(int[] nums, int k) {Map<Integer, Integer> freqs = new HashMap<>();for (int num : nums) {freqs.put(num, freqs.getOrDefault(num, 0) + 1);}// 1.PQPriorityQueue<Map.Entry<Integer, Integer>> minHeap = new PriorityQueue<>(new EntryComparator());for (Map.Entry<Integer, Integer> e : freqs.entrySet()) {minHeap.offer(e);if (minHeap.size() > k) {minHeap.poll();}}List<Integer> ans = new ArrayList<>();for (int i = 0; i < k; i++) {// P1:注意加到答案的是key不是entry
            ans.add(minHeap.poll().getKey());}Collections.reverse(ans);return ans;}
}

class Solution {public List<Integer> topKFrequent(int[] nums, int k) {Map<Integer, Integer> freqs = new HashMap<>();for (int num : nums) {freqs.put(num, freqs.getOrDefault(num, 0) + 1);}// 1.bucket sort// P1: 注意这个初始化！List<Integer>[] buckets = new List[nums.length + 1];for (int num : freqs.keySet()) {int frequency = freqs.get(num);if (buckets[frequency] == null) {buckets[frequency] = new ArrayList<>();}buckets[frequency].add(num);}List<Integer> ans = new ArrayList<>();for (int i = buckets.length - 1; i >= 0; i--) {if (buckets[i] == null) {continue;}for (int j = 0; j < buckets[i].size() && k > 0; j++) {ans.add(buckets[i].get(j));k--;}}return ans;}
}