heapq模块拥有实现maxheap所需的一切。 它只做max-heap的堆推功能。 我已在下面示范如何克服这一点

在heapq模块中添加这个函数:

def _heappush_max(heap, item):
    """Push item onto heap, maintaining the heap invariant."""
    heap.append(item)
    _siftdown_max(heap, 0, len(heap)-1)

最后加上这句话:

try:
    from _heapq import _heappush_max
except ImportError:
    pass

瞧!这是完成了。

PS -转到heapq函数。首先在编辑器中写入“import heapq”，然后右键单击“heapq”并选择转到定义。

最简单的方法是反转键的值并使用heapq。例如，将1000.0转换为-1000.0，将5.0转换为-5.0。

如果插入的键具有可比性但不像int型，则可能重写它们上的比较操作符(即<=变成>，>变成<=)。否则，您可以重写heapq。heapq模块中的_siftup(最后都是Python代码)。

我创建了一个名为heap_class的包，它实现了最大堆，还将各种堆函数包装到一个与列表兼容的环境中。

>>> from heap_class import Heap
>>> h = Heap([3, 1, 9, 20], max=True)
>>> h.pop()
20
>>> h.peek()  # same as h[0]
9
>>> h.push(17)  # or h.append(17)
>>> h[0]  # same as h.peek()
17
>>> h[1]  # inefficient, but works
9

从最大堆中获得最小堆。

>>> y = reversed(h)
>>> y.peek()
1
>>> y  # repr is inefficient, but correct
Heap([1, 3, 9, 17], max=False)
>>> 9 in y
True
>>> y.raw()  # underlying heap structure
[1, 3, 17, 9]

正如其他人所提到的，在max堆中处理字符串和复杂对象在heapq中是相当困难的，因为它们不同 否定的形式。heap_class实现简单:

>>> h = Heap(('aa', 4), ('aa', 5), ('zz', 2), ('zz', 1), max=True)
>>> h.pop()
('zz', 2)

支持自定义键，并与后续的推/追加和弹出一起工作:

>>> vals = [('Adam', 'Smith'), ('Zeta', 'Jones')]
>>> h = Heap(vals, key=lambda name: name[1])
>>> h.peek()  # Jones comes before Smith
('Zeta', 'Jones')
>>> h.push(('Aaron', 'Allen'))
>>> h.peek()
('Aaron', 'Allen')

(实现是建立在heapq函数上的，所以它都是用C语言或C语言包装的，除了Python中max heap上的heappush和heapreplace)

扩展int类并重写__lt__是一种方法。

import queue
class MyInt(int):
    def __lt__(self, other):
        return self > other

def main():
    q = queue.PriorityQueue()
    q.put(MyInt(10))
    q.put(MyInt(5))
    q.put(MyInt(1))
    while not q.empty():
        print (q.get())


if __name__ == "__main__":
    main()

python中有内置堆，但我只是想分享一下，如果有人像我一样想自己构建它。 我是python的新手，不要判断我是否犯了错误。 算法是有效的，但效率我不知道

class Heap :

    def __init__(self):
        self.heap = []
        self.size = 0


    def add(self, heap):
        self.heap = heap
        self.size = len(self.heap)

    def heappush(self, value):
        self.heap.append(value)
        self.size += 1


    def heapify(self, heap ,index=0):

        mid = int(self.size /2)
        """
            if you want to travel great value from bottom to the top you need to repeat swaping by the hight of the tree
            I  don't how how can i get the  height of the tree that's why i use sezi/2
            you can find height by this formula
            2^(x) = size+1  why 2^x because tree is growing exponentially 
            xln(2) = ln(size+1)
            x = ln(size+1)/ln(2)
        """

        for i in range(mid):
            self.createTee(heap ,index)

        return heap

    def createTee(self,  heap ,shiftindex):

        """
        """
        """

            this pos reffer to the index of the parent only parent with children
                    (1)
                (2)      (3)           here the size of list is 7/2 = 3
            (4)   (5)  (6)  (7)        the number of parent is 3 but we use {2,1,0} in while loop
                                       that why a put pos -1

        """
        pos = int(self.size /2 ) -1
        """
            this if you wanna sort this heap list we should swap max value in the root of the tree with the last
            value in the list and if you wanna repeat this until sort all list you will need to prevent the func from
            change what we already sorted I should decrease the size of the list that will heapify on it

        """

        newsize = self.size - shiftindex
        while pos >= 0 :
            left_child = pos * 2 + 1
            right_child = pos * 2 + 2
            # this mean that left child is exist
            if left_child < newsize:
                if right_child < newsize:
                    # if the right child exit we wanna check if left child > rightchild
                    # if right child doesn't exist we can check that we will get error out of range
                    if heap[pos] < heap[left_child] and heap[left_child]  > heap[right_child] :
                        heap[left_child] , heap[pos] = heap[pos], heap[left_child]
                # here if the righ child doesn't exist
                else:
                    if heap[pos] < heap[left_child] :
                        heap[left_child] , heap[pos] = heap[pos], heap[left_child]
            # if the right child exist
            if right_child < newsize :
                if heap[pos] < heap[right_child] :
                    heap[right_child], heap[pos] = heap[pos], heap[right_child]
            pos -= 1

        return heap

    def sort(self ):
        k = 1
        for i in range(self.size -1 ,0 ,-1):
            """
            because this is max heap we swap root with last element in the list

            """
            self.heap [0] , self.heap[i] = self.heap[i], self.heap[0]
            self.heapify(self.heap ,k)
            k+=1

        return self.heap


h = Heap()
h.add([5,7,0,8,9,10,20,30,50,-1] )
h.heappush(-2)
print(" before heapify ")
print(h.heap)
print(" after heapify ")
print(h.heapify(h.heap,0))
print(" after sort ")
print(h.sort())

输出:

之前heapify [5,7,0,8,9,10,20,30,50， -1， -2]

heapify后 [50, 30, 20, 8, 9, 10, 0, 7, 5， -1， -2]

排序后 [-2， -1, 0,5,7,8,9,10,20,30,50]

希望您能理解我的代码。如果有什么你不明白的地方，请发表评论，我会尽力帮助你