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heapify函数及其作用。我在最大堆中看到了一个问题:
for i in range(n, -1, -1):
根据我的研究,我认为我只需要在非叶子节点上建立最大堆,应该为0 ... n / 2。所以这里的范围正确吗?
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 ... 0到step = -1,这个范围如何工作?
答案 0 :(得分:0)
用于C ++的HeapSort的GeeksforGeeks代码
// Build heap (rearrange array)
for (int i = n / 2 - 1; i >= 0; i--)
heapify(arr, n, i);
参考:-
CLRS书
中的Heapsort伪代码BUILD-MAX-HEAP(A)
heap-size[A] ← length[A]
for i ← length[A]/2 downto 1
do MAX-HEAPIFY(A, i)
是的,您是对的。仅堆非叶节点就足够了。
关于第二个问题:-
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.