347. Top K Frequent Elements java solutions

Given a non-empty array of integers, return the k most frequent elements.

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].

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.

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 1 public class Solution {
 2     public List<Integer> topKFrequent(int[] nums, int k) {
 3         HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
 4         ArrayList<Integer>[] bucket = new ArrayList[nums.length+1];
 5         for(int n : nums){
 6             map.put(n,map.get(n) == null ? 1 : map.get(n)+1);
 7         }
 8         for(int key : map.keySet()){
 9             int cnt = map.get(key);
10             if(bucket[cnt] == null){
11                 bucket[cnt] = new ArrayList<>();
12             }
13             bucket[cnt].add(key);
14         }
15         List<Integer> ans = new ArrayList<Integer>();
16         for(int i = bucket.length-1;i>=0;i--){
17             if(bucket[i] != null){
18                 ans.addAll(bucket[i]);
19                 if(ans.size() >= k) break;
20             }
21         }
22         return ans;
23     }
24 }

最初的想法是使用hashmap 记录频率，之后再按频率大小逆序输出k个元素。之后参考 discuss 中的代码，使用bucket来降低时间复杂度达到O(n)。