数据结构之优先队列：最小索引优先队列 Python代码实现——15

最小索引优先队列(Min index priority queue)

在之前实现的最大优先队列和最小优先队列,他们可以分别快速访问到队列中最大元索和最小元素,但是他们有一个缺点,就是没有办法通过索引访问已存在于优先队列中的对象,并更新它们。

为了实现这个目的,在优先队列的基础上,学习一种新的数据结构,索引优先队列。

接下来我们以最小索引优先队列举列，最大优先索引队列，有兴趣可以自行实现。

实现思路

实现功能的三个重要数组

数组items，储存无序插入的元素的数组数组pq，假设将items储存的元素进行从小到大排序，并且items对应的原索引依旧保持不变，而pq储存的元素就是items排序过后元素的原索引（参照下图可更好地理解）数组qp，qp的元素对应pq中的索引， qp的索引对应pq中的元素（也就对应着未排序的items中的元素，与items元素的索引保持一致）

这样删除时只需要根据传入的索引在qp中寻找对应的元素就可以跟快地找到了

实现的操作方法

size()获取队列的大小is_empty()判断队列是否为空less(x, y)对传入的两个索引对应当前队列的元素进行大小比较swap(i, j)对传入的两个索引对应当前队列中的元素进行值交换min_elem_index()获取最小元素的索引is_index_exist()判断索引在当前队列中是否对应一个元素insert(index, item)指定索引index处插入一个元素itemdelete_min_elem()删除最小元素，并返回最小元素插入队列时的索引change_item(idx, itm)指定索引idx处，将该处元素替换为itmswim()上浮排序操作，同之前堆的排序中介绍sink()下沉排序操作，同之前堆的排序中介绍

Python代码实现

import operatorclass IndexMinPriorityQueue:def __init__(self, length):self.items = [None for _ in range(length)]# Ascendingly sort items and memorize the item's relative index in itemsself.pq = [None] + [i if self.items[i] else None for i in range(len(self.items))]# Its index is associate with elements in pq, and also syncs with indices of list itemsself.qp = [i if self.pq[i] else None for i in range(len(self.pq))]self.N = 0def size(self):return self.Ndef is_empty(self):return self.N == 0def less(self, i, j):"""Compare the given two items in self.items"""return operator.lt(self.items[self.pq[i]], self.items[self.pq[j]])def swap(self, i, j):"""But change the position of the two items in the subsidiary pq and qp lists"""self.pq[i], self.pq[j] = self.pq[j], self.pq[i]self.qp[self.pq[i]], self.qp[self.pq[j]] = i, jdef min_elem_index(self):"""Find the minimum element's index"""return self.pq[1]def is_index_exist(self, index):"""Judge if the given index is exist in this queue"""return self.qp[index] is not Nonedef insert(self, index, item):"""Insert an element associated with the element's index in this queue"""if self.is_index_exist(index):returnself.items[index] = itemself.N += 1# Now it isn't a orderly queueself.pq[self.N] = indexself.qp[index] = self.N# swim the last element to make list pq orderedself.swim(self.N)def delete_min_elem(self):"""Delete the minimum element, and return its index"""min_index = self.pq[1]# print(f"min_ele: {self.items[min_index]}")self.swap(1, self.N)self.pq[self.N] = Noneself.qp[min_index] = Noneself.items[min_index] = Noneself.N -= 1self.sink(1)return min_indexdef change_item(self, idx, itm):"""Substitute a item which index=idx with a new item which value=itm"""self.items[idx] = itmk = self.qp[idx]self.sink(k)self.swim(k)def swim(self, index):"""Move the smaller element up; We should only change order in pq and qp"""while index > 1:# Compare the current node with its parent node, if smaller, swim upif self.less(index, int(index/2)): # Compare values in itemsself.swap(index, int(index/2)) # But swap the mapping position in pq and qpindex = int(index/2)def sink(self, index):"""Move the bigger element down; We should only change order in pq and qp"""# print(f"SINK: idx:{index} N:{self.N}")while 2*index <= self.N:index_smaller = 2*index if 2*index+1 > self.N else \(2*index if self.less(2*index, 2*index+1) else 2*index+1)# print(f"index_smaller: {index_smaller}")# print(f"index: {index}")if self.less(index, index_smaller):breakself.swap(index, index_smaller)index = index_smaller

测试代码

if __name__ == '__main__':IMPQ = IndexMinPriorityQueue(10)IMPQ.insert(0, 'C')IMPQ.insert(1, 'E')IMPQ.insert(2, 'A')print(f"After three insert list items now is: {IMPQ.items}")# print(f"pq: {IMPQ.pq}")# print(f"qp: {IMPQ.qp}")# print(f"min_elem_index: {IMPQ.min_elem_index()}")index, item = 0, 'B'IMPQ.change_item(0, 'B')print(f"Changed the item in index[{index}] to {item}, items now is {IMPQ.items}")index, item = 0, 'V'IMPQ.change_item(0, 'V')print(f"Changed the item in index[{index}] to {item}, items now is {IMPQ.items}")while not IMPQ.is_empty():res = IMPQ.delete_min_elem()print(f"Pop the minimum element: {res}")print(f"After delete all elements , its size: {IMPQ.size()}")print(IMPQ.is_index_exist(0))

测试结果

After three insert list items now is: ['C', 'E', 'A', None, None, None, None, None, None, None]Changed the item in index[0] to B, items now is ['B', 'E', 'A', None, None, None, None, None, None, None]Changed the item in index[0] to V, items now is ['V', 'E', 'A', None, None, None, None, None, None, None]Pop the minimum element: 2Pop the minimum element: 1Pop the minimum element: 0After delete all elements , its size: 0False

这里为了测试结果直观，直接将数组items拿出来了，注意items中的元素是不会改变位置的，实际改变的是用来定位items中元素的pq和qp数组