帮我写一段代码完成一个单链表的堆排序

好的，以下是一个单链表堆排序的示例代码：

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def heap_sort(head):
    if not head:
        return None
    heap = []
    cur = head
    while cur:
        heap.append(cur.val)
        cur = cur.next
    n = len(heap)
    # 建立小根堆
    for i in range(n // 2 - 1, -1, -1):
        heapify(heap, n, i)
    # 堆排序
    for i in range(n - 1, 0, -1):
        heap[0], heap[i] = heap[i], heap[0]
        heapify(heap, i, 0)
    # 构建排序后的链表
    dummy = ListNode(0)
    cur = dummy
    for val in heap:
        cur.next = ListNode(val)
        cur = cur.next
    return dummy.next

def heapify(heap, n, i):
    smallest = i
    left = 2 * i + 1
    right = 2 * i + 2
    if left < n and heap[left] < heap[smallest]:
        smallest = left
    if right < n and heap[right] < heap[smallest]:
        smallest = right
    if smallest != i:
        heap[i], heap[smallest] = heap[smallest], heap[i]
        heapify(heap, n, smallest)

使用方法：

l = ListNode(4, ListNode(2, ListNode(1, ListNode(3))))
l_sorted = heap_sort(l)

这个算法的时间复杂度为 O(nlogn)，其中 n 是链表的长度。