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如何将链表复制到另一个列表?

[英]How do you copy a linked list into another list?

I'm studying data structures and linked lists, but I'm not getting the concept of how to make a copy of a linked list. 我正在研究数据结构和链表,但我没有得到如何制作链表副本的概念。 Can someone explain this, possibly using pseudocode or C code? 有人可以解释一下,可能使用伪代码或C代码吗?

The logic for duplicating a linked list is recursive and based on the following observations: 复制链表的逻辑是递归的,并基于以下观察:

  1. The clone of the empty list is the empty list. 空列表的克隆是空列表。
  2. The clone of a list with first node x and remaining nodes xs is a copy of x prepended to a clone of xs. 具有第一节点x和剩余节点xs的列表的克隆是x的副本,其前缀为xs的克隆。

If you encode the linked list in C++, this can be very clean: 如果用C ++编码链表,这可能非常干净:

struct Node {
    int value;
    Node* next;
};

Node* Clone(Node* list) {
    if (list == NULL) return NULL;

    Node* result = new Node;
    result->value = list->value;
    result->next = Clone(list->next);
    return result;
}

Do you understand how to add a new node to an existing list? 您是否了解如何将新节点添加到现有列表? And do you understand how to traverse (ie iterate over) a list? 您是否了解如何遍历(即迭代)列表? Copying a list is just performing both of these operations simultaneously (traverse ListA; for each element, copy the element and add it as a new node to ListB). 复制列表只是同时执行这两个操作(遍历ListA;对于每个元素,复制元素并将其作为新节点添加到ListB)。

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