Python 中内置的最大堆 API

2023-09-29Python开发问题
5

本文介绍了Python 中内置的最大堆 API的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

默认 heapq 是最小队列实现,想知道是否有最大队列选项?谢谢.

Default heapq is min queue implementation and wondering if there is an option for max queue? Thanks.

我尝试了使用 _heapify_max 作为最大堆的解决方案,但是如何动态处理 push/pop 元素?看来 _heapify_max 只能在初始化期间使用.

I tried the solution using _heapify_max for max heap, but how to handle dynamically push/pop element? It seems _heapify_max could only be used during initialization time.

import heapq

def heapsort(iterable):
    h = []
    for value in iterable:
        heapq.heappush(h, value)
    return [heapq.heappop(h) for i in range(len(h))]

if __name__ == "__main__":

    print heapsort([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])

编辑,尝试 _heapify_max 似乎不适用于动态推送/弹出元素.我试过两种方法输出一样,输出都是,[0, 1, 2, 3, 4, 5, 6, 7, 8, 9].

Edit, tried _heapify_max seems not working for dynamically push/pop elements. I tried both methods output the same, both output is, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9].

def heapsort(iterable):
    h = []
    for value in iterable:
        heapq.heappush(h, value)
    return [heapq.heappop(h) for i in range(len(h))]

def heapsort2(iterable):
    h = []
    heapq._heapify_max(h)
    for value in iterable:
        heapq.heappush(h, value)
    return [heapq.heappop(h) for i in range(len(h))]

if __name__ == "__main__":

    print heapsort([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])
    print heapsort2([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])

提前致谢,林

推荐答案

过去我只是简单地使用 sortedcontainers 的 SortedList 为此,如:

In the past I have simply used sortedcontainers's SortedList for this, as:

> a = SortedList()
> a.add(3)
> a.add(2)
> a.add(1)
> a.pop()
3

它不是堆,但速度很快,可以根据需要直接工作.

It's not a heap, but it's fast and works directly as required.

如果你绝对需要它成为一个堆,你可以创建一个通用的否定类来保存你的项目.

If you absolutely need it to be a heap, you could make a general negation class to hold your items.

class Neg():
    def __init__(self, x):
        self.x = x

    def __cmp__(self, other):
        return -cmp(self.x, other.x)

def maxheappush(heap, item):
    heapq.heappush(heap, Neg(item))

def maxheappop(heap):
    return heapq.heappop(heap).x

但这会占用更多内存.

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