原题
Given a non-empty list of words, return the k most frequent elements.
Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.
Example 1:
Input: [“i”, “love”, “leetcode”, “i”, “love”, “coding”], k = 2
Output: [“i”, “love”]
Explanation: “i” and “love” are the two most frequent words.
Note that “i” comes before “love” due to a lower alphabetical order.
Example 2:
Input: [“the”, “day”, “is”, “sunny”, “the”, “the”, “the”, “sunny”, “is”, “is”], k = 4
Output: [“the”, “is”, “sunny”, “day”]
Explanation: “the”, “is”, “sunny” and “day” are the four most frequent words,
with the number of occurrence being 4, 3, 2 and 1 respectively.
Note:
You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Input words contain only lowercase letters.
Follow up:
Try to solve it in O(n log k) time and O(n) extra space.
解法
使用collections.counter()统计单词频率, 然后使用sorted()对字典的键进行排序, 先按照频率倒序排列, 再按照字母顺序排列, 最后返回前k个键.
Time: O(n*logk)
Space: O(n)
代码
class Solution(object):
def topKFrequent(self, words, k):
"""
:type words: List[str]
:type k: int
:rtype: List[str]
"""
d = collections.Counter(words)
res = sorted(d, key = lambda word: (-d[word], word))
return res[:k]
本文介绍了一种统计给定单词列表中出现频率最高的k个单词的方法。该方法首先利用collections.counter统计单词频率,然后通过sorted函数对统计结果进行排序,确保频率高的单词排在前面,相同频率的单词则按字母顺序排列。
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