输入一批正整数（数据个数未知，以0结束），用单链表存储，输出大于平均值的数。用Python编写

在Python中，你可以通过遍历链表并维护当前元素总和以及计数来解决这个问题。这里是一个基本的步骤：

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

def sum_and_count(head):
    total_sum = 0
    count = 0
    current = head
    while current and current.val != 0:
        total_sum += current.val
        count += 1
        current = current.next
    
    average = total_sum / count if count > 0 else 0  # 防止除数为零错误

    larger_than_average = []
    current = head
    while current:
        if current.val > average:
            larger_than_average.append(current.val)
        current = current.next
    
    return larger_than_average

# 示例：
# 定义链表节点
node1 = ListNode(5)
node2 = ListNode(2)
node3 = ListNode(8)
node4 = ListNode(1)
node5 = ListNode(9)  # 0 结束
node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5

result = sum_and_count(node1)
print("大于平均值的数有:", result)

这个函数首先计算出链表中所有数字的总和和个数，然后计算平均值。接着再次遍历链表，将大于平均值的数值添加到结果列表中。