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.

第一想法自然是放hash里面统计出现次数然后根据次数排序后取出前k个key

import operator
class Solution(object):
    def topKFrequent(self, nums, k):
        h = {}
        for x in nums:
            if h.has_key(x):
                h[x] += 1
            else:
                h[x] = 1
        sorted_h = sorted(h.items(), key=operator.itemgetter(1))
        sorted_h.reverse()
        return [x for (x,y) in sorted_h[:k]]