LeetCode : 347. Top K Frequent Elements 根据频率选取topk元素

试题：
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 n is the array’s size.
代码：
堆排序，先统计频率，然后维持一个堆。

class Solution {
  public List topKFrequent(int[] nums, int k) {
    HashMap counter = new HashMap<>();
    for (int num: nums) {
      counter.put(num, counter.getOrDefault(num, 0)+1);
    }

    PriorityQueue heap =
            new PriorityQueue((n1, n2) -> counter.get(n1) - counter.get(n2));

    for (int num: counter.keySet()) {
      heap.add(num);
      if (heap.size() > k)
        heap.poll();
    }

    List top_k = new LinkedList();
    while (!heap.isEmpty())
      top_k.add(heap.poll());
    Collections.reverse(top_k);
    return top_k;
  }
}

桶排序

class Solution {
    public List topKFrequent(int[] nums, int k) {
        Map counter = new HashMap<>();
        for(int num: nums){
            counter.put(num,counter.getOrDefault(num,0)+1);
        }
        
        List[] bucket = new ArrayList[nums.length+1];
        for(int key: counter.keySet()){
            int count = counter.get(key);
            if(bucket[count]==null){
                bucket[count] = new ArrayList<>();
            }
            bucket[count].add(key);
        }
        
        List topk = new ArrayList();
        for(int i=bucket.length-1; i>0&&topk.size()