Creating a linked list without pointers but indices [duplicate]

问题

My teacher wants me to create something like linked list but using indices instead of pointers. So the Node contains int data and int index.

Can you drop me a hint how would I do that? I know how to create it with pointers but how to do without them? He mentioned pool as a container though.

回答1:

Your struct Node would be like this

struct Node {
    int data;
    int index; // index of the entry in the pool to the next node
}

You can use a vector or array to create a pool

vector<Node*> pool; // stores the pointer to next nodes

Now to access the next node you can do

Node* nextNode = pool[currNode->index];

回答2:

A possible approach would be to use an array of Nodes where each node stores an (array) index to the prev and next Node. Thus, your Node would look something like:

struct Node 
{
    T data;
    int prev;    // index of the previous Node of the list
    int next;    // index of the next Node of the list
}

Additionally, you will probably have to dynamically allocate the Node array, implement some bookkeeping to get and free space in the array: for example a bool array that stores the unoccupied indexes in the Node array, along with two functions that will update it every time a new Node / index is added or removed (it will be fragmented as nodes won't be always contiguous); find an index of a Node in the array: for example by subtracting the address of the Node from the first address of the array.

Here is how a possible interface of a doubly linked list using the above technique would look like:

template <typename T>                          // T type Node in your case
class DList
{
    T** head;                                  // array of pointers of type Node
    int first;                                 // index of first Node
    int last;                                  // index of last Node

    bool* available;                           // array of available indexes 
    int list_size;                             // size of list

    int get_index();                           // search for index with value true in bool available
    void return_index(int i);                  // mark index as free in bool available

    std::ptrdiff_t link_index(T*& l) const     // get index of Node

    void init();                               // initialize data members
    void create();                             // allocate arrays: head and available

    void clear();                              // delete array elements
    void destroy();                            // delete head

public:
    DList();                                   // call create() and init()
    ~DList();                                  // call clear() and destroy()

    void push_back(T* l);
    void push_front(T* l);
    void insert(T*& ref, T* l);                // insert l before ref

    T* erase(T* l);                            // erase current (i.e. l) and return next
    T* advance(T* l, int n);                   // return n-th link before or after currnet

    std::size_t size() const;
    int capacity () const { return list_size; }
};

You could use that as a benchmark and implement something on your own.

template <typename T>
void DList<T>::push_back(T* l)
{
    if (l == nullptr)
    {
        throw std::invalid_argument("Dlist::push_back()::Null pointer as argument!\n");
    }

    int index = get_index();
    head[index] = l;

    if (last != -1)
    {
        head[last]->next = index;
        head[index]->prev = last;
    }
    else
    {
        first = index;
        head[index]->prev = -1;
    }

    last = index;
    head[index]->next = -1;
}

来源：https://stackoverflow.com/questions/44859479/creating-a-linked-list-without-pointers-but-indices

标签

c++

data-structures

linked-list