允许您选择任意数量的最大或最小的项目

import heapq
heap = [23, 7, -4, 18, 23, 42, 37, 2, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
heapq.heapify(heap)
print(heapq.nlargest(3, heap))  # [42, 42, 37]
print(heapq.nsmallest(3, heap)) # [-4, -4, 2]

我创建了一个堆包装器，它将值颠倒以创建max-heap，还为min-heap创建了一个包装器类，以使库更像oop。这是要点。有三个班级;Heap(抽象类)，HeapMin和HeapMax。

方法:

isempty() -> bool; obvious
getroot() -> int; returns min/max
push() -> None; equivalent to heapq.heappush
pop() -> int; equivalent to heapq.heappop
view_min()/view_max() -> int; alias for getroot()
pushpop() -> int; equivalent to heapq.pushpop

我还需要使用max-heap，我处理的是整数，所以我只是包装了我需要的heap中的两个方法，如下所示:

import heapq


def heappush(heap, item):
    return heapq.heappush(heap, -item)


def heappop(heap):
    return -heapq.heappop(heap)

然后，我只是分别用heappush()和heappop()替换了我的heapq.heappush()和heappop()调用。

允许您选择任意数量的最大或最小的项目

import heapq
heap = [23, 7, -4, 18, 23, 42, 37, 2, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
heapq.heapify(heap)
print(heapq.nlargest(3, heap))  # [42, 42, 37]
print(heapq.nsmallest(3, heap)) # [-4, -4, 2]

为了详细说明https://stackoverflow.com/a/59311063/1328979，这里有一个针对一般情况的完整文档、注释和测试的Python 3实现。

from __future__ import annotations  # To allow "MinHeap.push -> MinHeap:"
from typing import Generic, List, Optional, TypeVar
from heapq import heapify, heappop, heappush, heapreplace


T = TypeVar('T')


class MinHeap(Generic[T]):
    '''
    MinHeap provides a nicer API around heapq's functionality.
    As it is a minimum heap, the first element of the heap is always the
    smallest.
    >>> h = MinHeap([3, 1, 4, 2])
    >>> h[0]
    1
    >>> h.peek()
    1
    >>> h.push(5)  # N.B.: the array isn't always fully sorted.
    [1, 2, 4, 3, 5]
    >>> h.pop()
    1
    >>> h.pop()
    2
    >>> h.pop()
    3
    >>> h.push(3).push(2)
    [2, 3, 4, 5]
    >>> h.replace(1)
    2
    >>> h
    [1, 3, 4, 5]
    '''
    def __init__(self, array: Optional[List[T]] = None):
        if array is None:
            array = []
        heapify(array)
        self.h = array
    def push(self, x: T) -> MinHeap:
        heappush(self.h, x)
        return self  # To allow chaining operations.
    def peek(self) -> T:
        return self.h[0]
    def pop(self) -> T:
        return heappop(self.h)
    def replace(self, x: T) -> T:
        return heapreplace(self.h, x)
    def __getitem__(self, i) -> T:
        return self.h[i]
    def __len__(self) -> int:
        return len(self.h)
    def __str__(self) -> str:
        return str(self.h)
    def __repr__(self) -> str:
        return str(self.h)


class Reverse(Generic[T]):
    '''
    Wrap around the provided object, reversing the comparison operators.
    >>> 1 < 2
    True
    >>> Reverse(1) < Reverse(2)
    False
    >>> Reverse(2) < Reverse(1)
    True
    >>> Reverse(1) <= Reverse(2)
    False
    >>> Reverse(2) <= Reverse(1)
    True
    >>> Reverse(2) <= Reverse(2)
    True
    >>> Reverse(1) == Reverse(1)
    True
    >>> Reverse(2) > Reverse(1)
    False
    >>> Reverse(1) > Reverse(2)
    True
    >>> Reverse(2) >= Reverse(1)
    False
    >>> Reverse(1) >= Reverse(2)
    True
    >>> Reverse(1)
    1
    '''
    def __init__(self, x: T) -> None:
        self.x = x
    def __lt__(self, other: Reverse) -> bool:
        return other.x.__lt__(self.x)
    def __le__(self, other: Reverse) -> bool:
        return other.x.__le__(self.x)
    def __eq__(self, other) -> bool:
        return self.x == other.x
    def __ne__(self, other: Reverse) -> bool:
        return other.x.__ne__(self.x)
    def __ge__(self, other: Reverse) -> bool:
        return other.x.__ge__(self.x)
    def __gt__(self, other: Reverse) -> bool:
        return other.x.__gt__(self.x)
    def __str__(self):
        return str(self.x)
    def __repr__(self):
        return str(self.x)


class MaxHeap(MinHeap):
    '''
    MaxHeap provides an implement of a maximum-heap, as heapq does not provide
    it. As it is a maximum heap, the first element of the heap is always the
    largest. It achieves this by wrapping around elements with Reverse,
    which reverses the comparison operations used by heapq.
    >>> h = MaxHeap([3, 1, 4, 2])
    >>> h[0]
    4
    >>> h.peek()
    4
    >>> h.push(5)  # N.B.: the array isn't always fully sorted.
    [5, 4, 3, 1, 2]
    >>> h.pop()
    5
    >>> h.pop()
    4
    >>> h.pop()
    3
    >>> h.pop()
    2
    >>> h.push(3).push(2).push(4)
    [4, 3, 2, 1]
    >>> h.replace(1)
    4
    >>> h
    [3, 1, 2, 1]
    '''
    def __init__(self, array: Optional[List[T]] = None):
        if array is not None:
            array = [Reverse(x) for x in array]  # Wrap with Reverse.
        super().__init__(array)
    def push(self, x: T) -> MaxHeap:
        super().push(Reverse(x))
        return self
    def peek(self) -> T:
        return super().peek().x
    def pop(self) -> T:
        return super().pop().x
    def replace(self, x: T) -> T:
        return super().replace(Reverse(x)).x


if __name__ == '__main__':
    import doctest
    doctest.testmod()

https://gist.github.com/marccarre/577a55850998da02af3d4b7b98152cf4

最简单最理想的解决方案

将这些值乘以-1

好了。所有最高的数字现在都是最低的，反之亦然。

只要记住，当您弹出一个元素与-1相乘以再次获得原始值时。