[leetcode]347. Top K Frequent Elements K个最常见元素

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.

题目

给定数组，求其中出现频率最高的K个元素。

思路

bucket sort

代码

 /*
 Time: O(n)
 Space: O(n)
 */
 class Solution {
      public List<Integer> topKFrequent(int[] nums, int k) {
         // freq map
         Map<Integer, Integer> map = new HashMap<>();
         for (int num : nums) {
             map.put(num, map.getOrDefault(num, 0) + 1);
         }
         // bucket sort on freq
         List<Integer>[] buckets = new List[nums.length + 1];
         for (int i : map.keySet()) {
             int freq = map.get(i);
             if (buckets[freq] == null) {
                 buckets[freq] = new ArrayList<>();
             }
             buckets[freq].add(i);
         }
         // gather result
         List<Integer> res = new ArrayList<>();
         for (int i = buckets.length - 1; i >= 0; --i) {
             if (buckets[i] == null) continue;
             for (int item : buckets[i]) {
                 res.add(item);

                 if (k == res.size()) return res;
             }
         }
         return res;
     }
 }

思路

priorityqueue to track Top K Frequent Elements

1. Using HashMap and PriorityQueue
2. Build a HashMap to get the item frequency
3. PriorityQueue (minHeap) 
4.  If the PQ size > k, then we pop the lowest value in the PQ

跟[leetcode]692. Top K Frequent Words K个最常见单词完全思路一致

代码

 /*
 Time: O(nlogk)
 Space: O(k)
 */
 class Solution {
     public List<Integer> topKFrequent(int[] nums, int k) {
         // freq map
         Map<Integer, Integer> map = new HashMap();
         for(int num : nums){
             map.put(num, map.containsKey(num) ? map.get(num) + 1 : 1);
         }
         // maintain a k-size minHeap
         PriorityQueue <Map.Entry<Integer, Integer>> minHeap = new PriorityQueue<>((entry1, entry2) -> entry2.getValue() - entry1.getValue());

         for(Map.Entry<Integer, Integer> entry : map.entrySet()){
             minHeap.add(entry);
         }

         List<Integer> result = new ArrayList<>();
         while(!minHeap.isEmpty() && result.size() < k){
             result.add(0, minHeap.remove().getKey());
         }
         return result;
     }
 }