Question

以下是我对Heap Sort的尝试，它假设类似于第152页及以后的CLRS中显示的内容。

如果我传递A = [9,0,5,7,4,6,3,8,1,2]作为输入。 BuildMaxHeap的输出是[9,8,6,7,4,5,3,0,1,2]。这似乎是对的。然而，将相同的输入传递给HeapSort给了我[9,8,4,7,2,3,5,6,1,0]，这是错误的。

任何人都可以对我做错了吗？

def left(i):
    return 2 * i

def right(i):
    return 2 * i + 1

def HeapSize(A):
    return len(A) - 1 

def MaxHeapify(A, i): 
    l = left(i)
    r = right(i)
    if l <= HeapSize(A) and A[l] > A[i]:
            largest = l 
    else:
            largest = i 
    if r <= HeapSize(A) and A[r] > A[largest]:
            largest = r 
    if largest != i:
            A[i], A[largest] = A[largest], A[i]
            MaxHeapify(A, largest)

def BuildMaxHeap(A):
    for i in range(HeapSize(A) // 2, 0, -1):
            MaxHeapify(A, i)

def HeapSort(A):
    BuildMaxHeap(A)
    for i in range(HeapSize(A), 1, -1):
            A[i], A[1] = A[1], A[i]
            MaxHeapify(A, 1)

Answer 1

正如我所看到的，对原始代码进行了一些更正（查看评论）：

def left(i):
    return 2 * i      # this is 1-based tree. Since your first element is stored in 0th cell, here is an error

def right(i):
    return 2 * i + 1    # same, this's 1-based

def HeapSize(A):
    return len(A) - 1 

def MaxHeapify(A, i): 
    l = left(i)
    r = right(i)
    if l <= HeapSize(A) and A[l] > A[i]:
            largest = l       # always keep 4 white space (or tab) as indent in Python, don't mix up
    else:
            largest = i 
    if r <= HeapSize(A) and A[r] > A[largest]:
            largest = r 
    if largest != i:
            A[i], A[largest] = A[largest], A[i]
            MaxHeapify(A, largest)

def BuildMaxHeap(A):
    for i in range(HeapSize(A) // 2, 0, -1):     # the last non-leaf element is (HeapSize(A)-1) // 2
            MaxHeapify(A, i)

def HeapSort(A):
    BuildMaxHeap(A)
    for i in range(HeapSize(A), 1, -1):   # e.g. range(3,1,-1) gives you [3,2], you miss the 1st element
            A[i], A[1] = A[1], A[i]     # swap with A[0], unless you don't use the 0th element of array
            MaxHeapify(A, 1)      # here is essential error for the algorithm. Since you're doing in-place sorting, you'd not swap the largest element to end of heap then MaxHeapify the WHOLE ARRAY again. I.e. you have to MaxHeapify the heap EXCLUDING the last element

以下是基于您的代码的工作版本：

def left(i):
    return 2 * i + 1

def right(i):
    return 2 * i + 2

def parent(i):     # your 'left' and 'right' function is good practice, keep it
    return (i - 1) // 2

def MaxHeapify(A, heap_size, i): 
    l = left(i)
    r = right(i)
    if l <= heap_size and A[l] > A[i]:
        largest = l 
    else:
        largest = i 
    if r <= heap_size and A[r] > A[largest]:
        largest = r 
    if largest != i:
        A[i], A[largest] = A[largest], A[i]
        MaxHeapify(A, heap_size, largest)      # important: don't always sort the whole array (size of heap keeps decreasing in 'HeapSort')

def BuildMaxHeap(A):
    heap_size = len(A) - 1    # index of last element of the heap
    for i in range(parent(heap_size), -1, -1):    # don't miss i=0 iteration
        MaxHeapify(A, heap_size, i)

def HeapSort(A):
    BuildMaxHeap(A)
    heap_size = len(A) - 1
    for i in range(heap_size, 0, -1):    # ends at i=1. You don't swap A[0] with A[0], right?
        A[i], A[0] = A[0], A[i]
        MaxHeapify(A, i-1, 0)     # careful here: MaxHeapify which part of the array

测试结果：

>>> a = [9, 0, 5, 7, 4, 6, 3, 8, 1, 2]
>>> BuildMaxHeap(a)
>>> a
[9, 8, 6, 7, 4, 5, 3, 0, 1, 2]
>>> HeapSort(a)
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

越野车堆排序

1 个答案: