Data Structure for phone book [duplicate]

Answer 1

You Should use TRIE data Structure for Implementing Phonebook. TRIE is an ordered tree data structure that uses strings as keys. Unlike Binary Trees, TRIE does not store keys associated with the node.

Good example

Answer 2

in a phone book, the telephone number should be unique, address is unique, but the name could be duplicated.

so perhaps you can use hash table combine with linked list to approach this.

you can use any one or combination of the 'name, address, phone number' as hash key, if you simply use name as hash key, then linked list is needed to store the duplicated entries.

in this approach, search based on the hash key is O(1) efficiency, but search based on the other two will be O(n).

Answer 3

In C, I think a struct is the best option.

typedef struct _Contact Contact;

struct _Contact
{
    char* name;
    char* number;
    char* address;
};

Contact* add_new_contact( char* name, char* number, char* address )
{
    Contact* c = (Contact*) malloc( sizeof( Contact ) );
    c->name = name;
    c->number = number;
    c->address = address;
    return c;
}

Contact* phone_book [ 20 ];  /* An array of Contacts */

Use the standard string functions ( <string.h> or if using a C++ compiler, <cstring> ) or something like the glib for searching the names, numbers etc.

Here's a simple example:

Contact* search_for_number( Contact* phone_book[], const char* number )
{
    register int i;
    for( i = 0; i < sizeof( phone_book ); i++)
    {
       if ( strcmp( phone_book[i]->number, number ) == 0 ) return phone_book[i];
    }
    return NULL;
}

There is also a good, helpful code example over here.

Alternatively

You may be able to use linked lists, but since C or the C standard library doesn't provide linked-lists, you either need to implement it yourself, or to use a third-party library.

I suggest using the g_linked_list in the glib.

Answer 4

C or C++ or C#?

Use a list of classes

    public class PhoneBook
{
    public string name;
    public string phoneNumber;
    public string address;
}

place this in a list and you have a phone book

Answer 5

You could accomplish this with a single hash table or other type of associative array (if you wanted to). For each person, just have three keys in the table (name, address, phone) all pointing to the same record.

Answer 6

I think a combination of a trie (each phone book entry is one leaf) and two skip lists (one for each name and address) could turn out to be effective.

Just assign each node one set of pointers to move along the name axis, and one set of pointers to move along the address axis (that is, to traverse the skip lists).