Python heapq LeetCode 692. Top K Frequent Words

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本文深入探讨了Python中heapq模块的使用方法，包括如何利用heapq.heappush和heapq.heappop进行堆操作，以及如何通过heapq.heapify将列表转换为堆。此外，还介绍了一个LeetCode题目示例，展示了如何自定义类并实现比较方法以满足特定需求。

默认最小在最前，这个和sorted一样，没有reverse，但是可以用tuple来玩

heapq是module（文件），所以import

heapq.heappush是直接调用这个module最外层的函数

import heapq
h1 = []

heapq.heappush(h1, (5, 'write code'))
heapq.heappush(h1, (7, 'release product'))
heapq.heappush(h1, (1, 'write spec'))
heapq.heappush(h1, (3, 'create tests'))
print(heapq.heappop(h1))
#(1, 'write spec')

h2 = [(7,'7'),(6,'6'),(5,'5'),(4,'4'),(3,'3'),(2,'2'),(1,'1')]
heapq.heapify(h2)
print(h2)
#[(1, '1'), (3, '3'), (2, '2'), (4, '4'), (6, '6'), (7, '7'), (5, '5')]

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来个LeetCode题目玩一玩，涉及到Python的重载：

from collections import Counter
import heapq

class FreqWord:
    def __init__(self, word, freq):
        self.word, self.freq = word, freq

    def __lt__(self, other):
        if (self.freq != other.freq):
            return self.freq < other.freq
        else:
            return self.word > other.word

class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        cw = Counter(words)
        heap = []
        for word, freq in cw.items():
            heapq.heappush(heap, FreqWord(word, freq))
            if len(heap) > k:
                retire = heapq.heappop(heap)
        heap.sort(reverse=True)
        return [x.word for x in heap]