Doubly-linked list/Element definition
You are encouraged to solve this task according to the task description, using any language you may know.
Define the data structure for a doubly-linked list element. The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list. The pointers should be mutable.
Ada
<ada> type Link; type Link_Access is access Link; type Link is record
Next : Link_Access := null; Prev : Link_Access := null; Data : Integer;
end record; </ada> In Ada 2005 this example can be written without declaration of an access type: <ada> type Link is limited record
Next : not null access Link := Link'Unchecked_Access; Prev : not null access Link := Link'Unchecked_Access; Data : Integer;
end record; </ada> Here the list element is created already pointing to itself, so that no further initialization is required. The type of the element is marked as limited indicating that such elements have referential semantics and cannot be copied.
Ada's standard container library includes a generic doubly linked list. The structure of the link element is private.
ALGOL 68
MODE LINK = STRUCT ( REF LINK next, REF LINK prev, INT data ); LINK example; ~
C
struct link ( struct link *next; struct link *prev; int data; );
Fortran
In ISO Fortran 95 or later:
type node real :: data type(node), pointer :: next => null(), previous => null() end type node ! ! . . . . ! type( node ), target :: head
Java
public class Node<T> { private T element; private Node<T> next, prev; public Node<T>(){ next = prev = element = null; } public Node<T>(Node<T> n, Node<T> p, T elem){ next = n; prev = p; element = elem; } public void setNext(Node<T> n){ next = n; } public Node<T> getNext(){ return next; } public void setElem(T elem){ element = elem; } public T getElem(){ return element; } public void setNext(Node<T> n){ next = n; } public Node<T> setPrev(Node<T> p){ prev = p; } public getPrev(){ return prev; } }
For use with Java 1.4 and below, delete all "<T>"s and replace T's with "Object".
OCaml
<ocaml>type 'a dlink = {
mutable data: 'a; mutable next: 'a dlink option; mutable prev: 'a dlink option;
}
let dlink_of_list li =
let rec aux prev_dlink = function | [] -> prev_dlink | hd::tl -> let dlink = { data = hd; prev = None; next = prev_dlink } in begin match prev_dlink with | None -> () | Some prev_dlink -> prev_dlink.prev <- Some dlink end; aux (Some dlink) tl in aux None (List.rev li)
let print_dlink f =
let rec aux = function | None -> () | Some{ data = d; prev = _; next = next } -> f d; aux next in aux
- </ocaml>
<ocaml># let dl = dlink_of_list [1;2;3;4;5] in
print_dlink (Printf.printf "%d\n") dl ;;
1 2 3 4 5 - : unit = ()</ocaml>
Pascal
type link_ptr = ^link; data_ptr = ^data; (* presumes that type 'data' is defined above *) link = record prev: link_ptr; next: link_ptr; data: data_ptr; end;
Perl
Just use an array. You can traverse and splice it any way. Linked lists are way too low level.
However, if all you have is an algorithm in another language, you can use references to accomplish the translation.
my %node = ( data => 'say what', next => \%foo_node, prev => \%bar_node, ); $node{next} = \%quux_node; # mutable
Pop11
uses objectclass; define :class Link; slot next = []; slot prev = []; slot data = []; enddefine;
Python
class Node(object): def __init__(self, data = None, prev = None, next = None): self.prev = prev self.next = next self.data = data def __str__(self): return str(self.data) def __repr__(self): return repr(self.data) def iter_forward(self): c = self while c != None: yield c c = c.next def iter_backward(self): c = self while c != None: yield c c = c.prev
Ruby
class ListNode attr_accessor :val, :nxt, :prv def initialize(mval,mprv=nil,mnxt=nil) self.val=mval self.prv=mprv prv.nxt=self if prv self.nxt=mnxt nxt.prv=self if nxt end def each(&b) yield val nxt.each(&b) if nxt self end include Enumerable end