优先队列的堆实现（python语言）

骑猪看日落 2022-05-23 09:54 253阅读 0赞

堆排序

# -*- coding: UTF-8 -*-
    
    
    class PrioQueue:  # 优先队列类
        def __init__(self, elist=[]):  # 新建
            self._elems = list(elist)
            if elist:
                self.buildheap()  # 新建堆
    
        def is_empty(self):  # 判空
            return not self._elems
    
        def peek(self):  # 查看
            if self.is_empty():
                raise PrioQueueError('in peek')  # 为空时返回错误
            return self._elems[0]  #
    
        def enqueue(self, e):  # 入队
            self._elems.append(None)  # 添加一个元素
            self.siftup(e, len(self._elems) - 1)  # 审查
    
        def siftup(self, e, last):  # 筛分
            elems, i, j = self._elems, last, (last - 1) // 2
            while i > 0 and e < elems[j]:
                elems[i] = elems[j]
                i, j = j, (j - 1) // 2
            elems[i] = e
    
        def dequue(self):  # 出队
            if self.is_empty():
                raise PrioQueueError('in peek')  # 为空时返回错误
            elems = self._elems
            e0 = elems[0]
            e = elems.pop()
            if len(elems):  # 审查
                self.siftdown(e, 0, len(elems))
            return e0
    
        def siftdown(self, e, begin, end):  # 筛分
            elems, i, j = self._elems, begin, begin * 2 + 1
            while j < end:
                if j + 1 < end and elems[j + 1] < elems[j]:
                    j += 1  # elems[j]不大于其兄弟节点的数据
                if e < elems[j]:  # e在三者中最小，已经找到其位置
                    break
                elems[i] = elems[j]  # elems[j]在三者中最小，上移
                i, j = j, 2 * j + 1
            elems[i] = e
    
        def buildheap(self):  # 建堆
            end = len(self._elems)
            for i in range(end // 2, -1, -1):
                self.siftdown(self._elems[i], i, end)
    
        def heap_sort(elems):  # 堆排序
            def siftdown(elem, e, begin, end):  # 审查
                i, j = begin, begin * 2 + 1
                while j < end:
                    if j + 1 < end and elems[j + 1] < elems[j]:
                        j += 1  # elems[j]不大于其兄弟节点的数据
                    if e < elems[j]:  # e在三者中最小，已经找到其位置
                        break
                    elems[i] = elems[j]  # elems[j]在三者中最小，上移
                    i, j = j, 2 * j + 1
                elems[i] = e
    
            end = len(elems)
            for i in range(end // 2, -1, -1):  # 建堆，从i开始以end为建堆范围的边界
                siftdown(elems, elems[i], i, end)
            for i in range((end - 1), 0, -1):  # 逐个取出最小元素，将其积累到标的最后，放一个退一步
                e = elems[i]
                elems[i] = elems[0]
                siftdown(elems, e, 0, i)