如何在类似结构的链表中插入新节点？另一个

class Node:
    def __init__(self, data=None):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def insert(self, data, position):
        new_node = Node(data)

        if position == 0:
            new_node.next = self.head
            self.head = new_node
        else:
            current = self.head
            for _ in range(position - 1):
                if current.next is None:
                    raise IndexError("Position out of range")
                current = current.next

            new_node.next = current.next
            current.next = new_node

    def display(self):
        current = self.head
        while current:
            print(current.data, end=" ")
            current = current.next
        print()

# 创建链表
linked_list = LinkedList()

# 插入节点
linked_list.insert(1, 0)  # 在位置0插入节点1
linked_list.insert(2, 1)  # 在位置1插入节点2
linked_list.insert(3, 2)  # 在位置2插入节点3

# 显示链表
linked_list.display()  # 输出：1 2 3

class Node:
    def __init__(self, data=None):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def delete(self, position):
        if self.head is None:
            raise IndexError("List is empty")

        if position == 0:
            self.head = self.head.next
        else:
            current = self.head
            for _ in range(position - 1):
                if current.next is None:
                    raise IndexError("Position out of range")
                current = current.next

            if current.next is None:
                raise IndexError("Position out of range")

            current.next = current.next.next

    def display(self):
        current = self.head
        while current:
            print(current.data, end=" ")
            current = current.next
        print()

# 创建链表
linked_list = LinkedList()

# 插入节点
linked_list.insert(1, 0)  # 在位置0插入节点1
linked_list.insert(2, 1)  # 在位置1插入节点2
linked_list.insert(3, 2)  # 在位置2插入节点3

# 删除节点
linked_list.delete(1)  # 删除位置1的节点

# 显示链表
linked_list.display()  # 输出：1 3