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。

 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

 

posted @ 2017-01-02 14:07  lettuan  阅读(109)  评论(0)    收藏  举报