Leetcode 347. Top K Frequent Elements

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.

Top K 问题的变形，可以用collections.Count产生hash table。然后根据key值（freq）进行heap sort，这里可以仅仅维护一个size为 k 的 heap。

 class Solution(object):

     def topKFrequent(self, nums, k):

         """

         :type nums: List[int]

         :type k: int

         :rtype: List[int]

         """

         counts = collections.Counter(nums)

         heap = []

         for key, cnt in counts.items():

             if len(heap) < k:

                 heapq.heappush(heap, (cnt, key))

             else:

                 if heap[0][0] < cnt:

                     heapq.heapreplace(heap, (cnt, key))

         return [x[1] for x in heap]

巴特西

Leetcode 347. Top K Frequent Elements

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