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

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

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()

如果你想用max heap得到最大的K元素，你可以做下面的技巧:

nums= [3,2,1,5,6,4]
k = 2  #k being the kth largest element you want to get
heapq.heapify(nums) 
temp = heapq.nlargest(k, nums)
return temp[-1]

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。