Python -- 堆数据结构 heapq

Python -- 堆数据结构 heapq - I love this game! - 博客频道 - CSDN.NET

Python -- 堆数据结构 heapq

分类： Python 2012-09-17 14:56 458人阅读 评论(0) 收藏 举报

python数据结构arraysalgorithmlistencoding

import heapq

help(heapq)

heapq 是一个最小堆，堆顶元素 a[0] 永远是最小的. 和 Java 中的优先队列类似.

-------------------------------------------------

Help on module heapq:

NAME
    heapq - Heap queue algorithm (a.k.a. priority queue).


FILE
    /usr/lib64/python2.4/heapq.py


DESCRIPTION
    Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
    all k, counting elements from 0.  For the sake of comparison,
    non-existing elements are considered to be infinite.  The interesting
    property of a heap is that a[0] is always its smallest element.
    
    Usage:
    
    heap = []            # creates an empty heap
    heappush(heap, item) # pushes a new item on the heap
    item = heappop(heap) # pops the smallest item from the heap
    item = heap[0]       # smallest item on the heap without popping it
    heapify(x)           # transforms list into a heap, in-place, in linear time
    item = heapreplace(heap, item) # pops and returns smallest item, and adds
                                   # new item; the heap size is unchanged
    
    Our API differs from textbook heap algorithms as follows:
    
    - We use 0-based indexing.  This makes the relationship between the
      index for a node and the indexes for its children slightly less
      obvious, but is more suitable since Python uses 0-based indexing.
    
    - Our heappop() method returns the smallest item, not the largest.
    
    These two make it possible to view the heap as a regular Python list
    without surprises: heap[0] is the smallest item, and heap.sort()
    maintains the heap invariant!


FUNCTIONS
    heapify(...)
        Transform list into a heap, in-place, in O(len(heap)) time.
    
    heappop(...)
        Pop the smallest item off the heap, maintaining the heap invariant.
    
    heappush(...)
        Push item onto heap, maintaining the heap invariant.
    
    heapreplace(...)
        Pop and return the current smallest value, and add the new item.
        
        This is more efficient than heappop() followed by heappush(), and can be
        more appropriate when using a fixed-size heap.  Note that the value
        returned may be larger than item!  That constrains reasonable uses of
        this routine unless written as part of a conditional replacement:
        
                if item > heap[0]:
                    item = heapreplace(heap, item)
    
    nlargest(...)
        Find the n largest elements in a dataset.
        
        Equivalent to:  sorted(iterable, reverse=True)[:n]
    
    nsmallest(...)
        Find the n smallest elements in a dataset.
        

        Equivalent to:  sorted(iterable)[:n]

构建元素个数为 K=5 的最小堆代码实例:

#!/usr/bin/env python
# -*- encoding: utf-8 -*-
# Author: kentzhan
#

import heapq
import random

heap = []
heapq.heapify(heap)
for i in range(15):
  item = random.randint(10, 100)
  print "comeing ", item,
  if len(heap) >= 5:
    top_item = heap[0] # smallest in heap
    if top_item < item: # min heap
      top_item = heapq.heappop(heap)
      print "pop", top_item,
      heapq.heappush(heap, item)
      print "push", item,
  else:
    heapq.heappush(heap, item)
    print "push", item,
  pass
  print heap
pass
print heap

print "sort"
heap.sort()

print heap

#!/usr/bin/env python
# -*- encoding: utf-8 -*-
# Author: kentzhan
#

import heapq
import random

heap = []
heapq.heapify(heap)
for i in range(15):
  item = random.randint(10, 100)
  print "comeing ", item,
  if len(heap) >= 5:
    top_item = heap[0] # smallest in heap
    if top_item < item: # min heap
      top_item = heapq.heappop(heap)
      print "pop", top_item,
      heapq.heappush(heap, item)
      print "push", item,
  else:
    heapq.heappush(heap, item)
    print "push", item,
  pass
  print heap
pass
print heap

print "sort"
heap.sort()

print heap

结果: