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。

1 class Solution(object): 2     def topKFrequent(self, nums, k): 3         """ 4         :type nums: List[int] 5         :type k: int 6         :rtype: List[int] 7         """ 8         counts = collections.Counter(nums) 9         heap = []10         for key, cnt in counts.items():11             if len(heap) < k:12                 heapq.heappush(heap, (cnt, key))13             else:14                 if heap[0][0] < cnt:15                     heapq.heapreplace(heap, (cnt, key))16         return [x[1] for x in heap]17