Python堆处理实现说明

Question

这是heapsort的python3实现，其中n是堆的大小。

def heapify(arr, n, i): 
    largest = i  
    l = 2 * i + 1     # left = 2*i + 1 
    r = 2 * i + 2     # right = 2*i + 2 

# See if left child of root exists and is 
# greater than root 
if l < n and arr[i] < arr[l]: 
    largest = l 

# See if right child of root exists and is 
# greater than root 
if r < n and arr[largest] < arr[r]: 
    largest = r 

# Change root, if needed 
if largest != i: 
    arr[i],arr[largest] = arr[largest],arr[i] # swap 

    # Heapify the root. 
    heapify(arr, n, largest) 

# The main function to sort an array of given size 
def heapSort(arr): 
   n = len(arr) 

   # Build a maxheap. 
   for i in range(n, -1, -1): 
       heapify(arr, n, i) 

# One by one extract elements 
for i in range(n-1, 0, -1): 
    arr[i], arr[0] = arr[0], arr[i] # swap 
    heapify(arr, i, 0) 

我了解保健功能和它在做什么。但我在max heap中发现了一个问题：

^{pr2}$

根据我的研究，我认为我需要建立最大堆只在非叶节点，这应该是0。。。n/2.那么这里的范围正确吗？

我也很难理解最后一部分：

for i in range(n-1, 0, -1): 
arr[i], arr[0] = arr[0], arr[i] # swap 
heapify(arr, i, 0)

从n-1开始这个范围是怎么工作的。。。0的步骤=-1？

Answer 1

c++堆排序的GeeksforGeeks代码

// Build heap (rearrange array) 
for (int i = n / 2 - 1; i >= 0; i ) 
    heapify(arr, n, i); 

参考文献：-

1. GeeksforGeeks

CLRS书籍中Heapsort的伪代码

^{pr2}$

所以是的，你是对的。只有不愈合的叶子才足够强壮。在

至于你的第二个呢问题：在

伪代码

1.MaxHeapify(Array)
2.So the Array[0] has the maximum element
3.Now exchange Array[0] and Array[n-1] and decrement the size of heap by 1. 
4.So we now have a heap of size n-1 and we again repeat the steps 1,2 and 3 till the index is 0.