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]