题目
给一非空的单词列表,返回前 k 个出现次数最多的单词。
返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率,按字母顺序排序。
示例 1:
输入: ["i", "love", "leetcode", "i", "love", "coding"]
, k = 2
输出: ["i", "love"]
解析: "i"
和 "love"
为出现次数最多的两个单词,均为2次。
注意,按字母顺序 "i"
在 "love"
之前。
示例 2:
输入: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"]
, k = 4
输出: ["the", "is", "sunny", "day"]
解析: "the"
, "is"
, "sunny"
和 "day"
是出现次数最多的四个单词,
出现次数依次为 4, 3, 2 和 1 次。
注意:
- 假定 k 总为有效值, 1 ≤ k ≤ 集合元素数。
- 输入的单词均由小写字母组成。
方案一(两次稳定的排序算法)
class Solution:
def topKFrequent(self, words: List[str], k: int) -> List[str]:
# 两次稳定的排序算法
words_count = []
words_mapping = {} # key: word, value: words_count_index
for word in words:
if word in words_mapping:
words_count[words_mapping[word]][1] += 1
else:
words_count.append([word, 1])
words_mapping[word] = len(words_count) - 1
words_count.sort(key=lambda t: t[0])
words_count.sort(key=lambda t: t[1], reverse=True)
return [word for word, count in words_count[:k]]
leetcode 写法
class Solution(object):
def topKFrequent(self, words, k):
count = collections.Counter(words)
candidates = count.keys()
candidates.sort(key = lambda w: (-count[w], w))
return candidates[:k]
方案二(堆)
class Solution(object):
def topKFrequent(self, words, k):
count = collections.Counter(words)
heap = [(-freq, word) for word, freq in count.items()]
heapq.heapify(heap)
return [heapq.heappop(heap)[1] for _ in xrange(k)]