347. Top K Frequent Elements (Medium)

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.

思路：利用collections.Counter的most_common方法

class Solution():
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        cnt = collections.Counter(nums)
        return [x[0] for x in cnt.most_common(k)]

347. Top K Frequent Elements (Medium)

原文：http://www.cnblogs.com/yancea/p/7581449.html