最好的方法:

from heapq import *
h = [5, 7, 9, 1, 3]
h_neg = [-i for i in h]
heapify(h_neg)            # heapify
heappush(h_neg, -2)       # push
print(-heappop(h_neg))    # pop
# 9

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

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

你可以使用

import heapq
listForTree = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]    
heapq.heapify(listForTree)             # for a min heap
heapq._heapify_max(listForTree)        # for a maxheap!!

如果你想要弹出元素，使用:

heapq.heappop(minheap)      # pop from minheap
heapq._heappop_max(maxheap) # pop from maxheap

解决方案是当你在堆中存储你的值时对其求反，或者像这样反转你的对象比较:

import heapq

class MaxHeapObj(object):
  def __init__(self, val): self.val = val
  def __lt__(self, other): return self.val > other.val
  def __eq__(self, other): return self.val == other.val
  def __str__(self): return str(self.val)

max-heap的例子:

maxh = []
heapq.heappush(maxh, MaxHeapObj(x))
x = maxh[0].val  # fetch max value
x = heapq.heappop(maxh).val  # pop max value

但是您必须记住包装和打开您的值，这需要知道您正在处理的是最小堆还是最大堆。

MinHeap, MaxHeap类

为MinHeap和MaxHeap对象添加类可以简化代码:

class MinHeap(object):
  def __init__(self): self.h = []
  def heappush(self, x): heapq.heappush(self.h, x)
  def heappop(self): return heapq.heappop(self.h)
  def __getitem__(self, i): return self.h[i]
  def __len__(self): return len(self.h)

class MaxHeap(MinHeap):
  def heappush(self, x): heapq.heappush(self.h, MaxHeapObj(x))
  def heappop(self): return heapq.heappop(self.h).val
  def __getitem__(self, i): return self.h[i].val

使用示例:

minh = MinHeap()
maxh = MaxHeap()
# add some values
minh.heappush(12)
maxh.heappush(12)
minh.heappush(4)
maxh.heappush(4)
# fetch "top" values
print(minh[0], maxh[0])  # "4 12"
# fetch and remove "top" values
print(minh.heappop(), maxh.heappop())  # "4 12"

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。