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

本文介绍了一种高效算法，用于从整数数组中找出出现频率最高的k个元素。通过堆排序和桶排序两种方法，实现时间复杂度优于O(nlogn)，适用于大数据集的快速处理。

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

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

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

    List<Integer> 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<Integer> topKFrequent(int[] nums, int k) {
        Map<Integer,Integer> counter = new HashMap<>();
        for(int num: nums){
            counter.put(num,counter.getOrDefault(num,0)+1);
        }
        
        List<Integer>[] 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<Integer> topk = new ArrayList<Integer>();
        for(int i=bucket.length-1; i>0&&topk.size()<k; i--){
            if(bucket[i]==null) continue;
            if(bucket[i].size()<=(k-topk.size())){
                topk.addAll(bucket[i]);
            }else{
                topk.addAll(bucket[i].subList(0,k-topk.size()));
            }
        }
        return topk;
    }
}