【python3】leetcode 347. Top K Frequent Elements（Medium）

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347. Top K Frequent Elements（Medium）

Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

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.

求频率最高的k个数字，时间复杂度要优于 O(n log n),

第一个想到先计算每个数字的频率-》collections.Counter

然后根据频率排序取最后k个的数字

做这道题卡在如何对字典排序，参考了《

(补充了一种直接使用count.most_common(n)的方法，）count.most_common(n)返回list，对count按照计数排序返回数量最多的n个，返回的list的格式是[(key1 , val1 ) , (key2,val2),...,(keyn,valn)]

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

下面两种都是把dict先排序 

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class Solution:
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        count = collections.Counter(nums)
        l = [[val,key] for key,val in count.items()]
        l.sort()
        freq =  [v[1] for v in l]
        return freq[-k:]

还有一种利用sorted()里的key参数根据dict的val排序，然后利用odereddict顺序不变的特性直接取末k个key返回 

class Solution:
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        count = collections.Counter(nums)
        freq = collections.OrderedDict(sorted(count.items(),key=lambda t:t[1])) #sort according to val
        freq = list(freq.keys())
        return freq[-k:]