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  • 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)]
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  • 原文地址:https://www.cnblogs.com/yancea/p/7581449.html
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