Python中的堆数据结构——heap模块

>>> import heapq>>> help(heapq)Help on module heapq:NAME    heapq - Heap queue algorithm (a.k.a. priority queue).FILE    c:\python27\lib\heapq.pyDESCRIPTION    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(...)        heappush(heap, item) -> None. Push item onto heap, maintaining the heap invariant.        heappushpop(...)        heappushpop(heap, item) -> value. Push item on the heap, then pop and return the smallest item        from the heap. The combined action runs more efficiently than        heappush() followed by a separate call to heappop().        heapreplace(...)        heapreplace(heap, item) -> value. 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)        merge(*iterables)        Merge multiple sorted inputs into a single sorted output.                Similar to sorted(itertools.chain(*iterables)) but returns a generator,        does not pull the data into memory all at once, and assumes that each of        the input streams is already sorted (smallest to largest).                >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))        [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]        nlargest(n, iterable, key=None)        Find the n largest elements in a dataset.                Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]        nsmallest(n, iterable, key=None)        Find the n smallest elements in a dataset.                Equivalent to:  sorted(iterable, key=key)[:n]DATA    __about__ = 'Heap queues\n\n[explanation by Fran\xe7ois Pinard]\n\nH.....    __all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge', '...

demo：

>>> a_list = [1,4,5,2,6,9,0,3]#把list转化成heap>>> heapq.heapify(a_list)   >>> a_list[0, 2, 1, 3, 6, 9, 5, 4]#选出n个最大和最小值。 O(n)时间复杂度>>> heapq.nlargest(3,a_list)[9, 6, 5]>>> heapq.nsmallest(3,a_list)[0, 1, 2]

>>> heapq.heappushpop(a_list,7)0#返回当前最小值，再添加新元素>>> heapq.heapreplace(a_list,-1)1>>> a_list[-1, 2, 5, 3, 6, 9, 7, 4]#添加新元素并返回最小值>>> heapq.heappushpop(a_list,-2)-2>>> a_list[-1, 2, 5, 3, 6, 9, 7, 4]