Doubly-linked list/Traversal
You are encouraged to solve this task according to the task description, using any language you may know.
Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
C
<lang c>// A doubly linked list of strings;
- include <stdio.h>
- include <stdlib.h>
- include <string.h>
typedef struct sListEntry {
const char *value; struct sListEntry *next; struct sListEntry *prev;
} *ListEntry, *LinkedList;
typedef struct sListIterator{
ListEntry link; LinkedList head;
} *LIterator;
LinkedList NewList() {
ListEntry le = (ListEntry)malloc(sizeof(struct sListEntry)); if (le) { le->value = NULL; le->next = le->prev = NULL; } return le;
}
int LL_Append(LinkedList ll, const char *newVal) {
ListEntry le = (ListEntry)malloc(sizeof(struct sListEntry)); if (le) { le->value = strdup(newVal); le->prev = ll->prev; le->next = NULL; if (le->prev) le->prev->next = le; else ll->next = le; ll->prev = le; } return (le!= NULL);
}
int LI_Insert(LIterator iter, const char *newVal) {
ListEntry crnt = iter->link; ListEntry le = (ListEntry)malloc(sizeof(struct sListEntry)); if (le) { le->value = strdup(newVal); if ( crnt == iter->head) { le->prev = NULL; le->next = crnt->next; crnt->next = le; if (le->next) le->next->prev = le; else crnt->prev = le; } else { le->prev = ( crnt == NULL)? iter->head->prev : crnt->prev; le->next = crnt; if (le->prev) le->prev->next = le; else iter->head->next = le; if (crnt) crnt->prev = le; else iter->head->prev = le; } } return (le!= NULL);
}
LIterator LL_GetIterator(LinkedList ll ) {
LIterator liter = (LIterator)malloc(sizeof(struct sListIterator)); liter->head = ll; liter->link = ll; return liter;
}
- define LLI_Delete( iter ) \
{free(iter); \ iter = NULL;}
int LLI_AtEnd(LIterator iter) {
return iter->link == NULL;
} const char *LLI_Value(LIterator iter) {
return (iter->link)? iter->link->value: NULL;
} int LLI_Next(LIterator iter) {
if (iter->link) iter->link = iter->link->next; return(iter->link != NULL);
} int LLI_Prev(LIterator iter) {
if (iter->link) iter->link = iter->link->prev; return(iter->link != NULL);
}
int main() {
static const char *contents[] = {"Read", "Orage", "Yeller", "Glean", "Blew", "Burple"}; int ix; LinkedList ll = NewList(); //new linked list LIterator iter;
for (ix=0; ix<6; ix++) //insert contents LL_Append(ll, contents[ix]);
iter = LL_GetIterator(ll); //get an iterator printf("forward\n"); while(LLI_Next(iter)) //iterate forward printf("value=%s\n", LLI_Value(iter)); LLI_Delete(iter); //delete iterator
printf("\nreverse\n"); iter = LL_GetIterator(ll); while(LLI_Prev(iter)) //iterate reverse printf("value=%s\n", LLI_Value(iter)); LLI_Delete(iter); //uhhh-- delete list?? return 0;
}</lang>
JavaScript
See Doubly-Linked List (element)#JavaScript. The traverse()
and print()
functions have been inherited from Singly-Linked List (traversal)#JavaScript.
<lang javascript>DoublyLinkedList.prototype.getTail = function() {
var tail; this.traverse(function(node){tail = node;}); return tail;
} DoublyLinkedList.prototype.traverseBackward = function(func) {
func(this); if (this.prev() != null) this.prev().traverseBackward(func);
} DoublyLinkedList.prototype.printBackward = function() {
this.traverseBackward( function(node) {print(node.value())} );
}
var head = createDoublyLinkedListFromArray([10,20,30,40]); head.print(); head.getTail().printBackward();</lang>
outputs:
10 20 30 40 40 30 20 10
Uses the print()
function from Rhino or SpiderMonkey.
Haskell
<lang haskell> main = print . traverse True $ create [10,20,30,40]
data DList a = Leaf | Node (DList a) a (DList a)
create = worker Leaf
where worker _ [] = Leaf worker prev (x:xs) = current where current = Node prev x next next = worker current xs
traverse _ Leaf = [] traverse True (Node l v Leaf) = v : v : traverse False l traverse dir (Node l v r) = v : traverse dir (if dir then r else l) </lang>
Oz
Warning: Highly unidiomatic code. (Use built-in lists instead.) <lang oz>declare
proc {Walk Node Action} case Node of nil then skip [] node(value:V next:N ...) then
{Action V} {Walk @N Action}
end end
proc {WalkBackwards Node Action} Tail = {GetLast Node} proc {Loop N}
case N of nil then skip [] node(value:V prev:P ...) then {Action V} {Loop @P} end
end in {Loop Tail} end
fun {GetLast Node} case @(Node.next) of nil then Node [] NextNode=node(...) then {GetLast NextNode} end end fun {CreateNewNode Value} node(prev:{NewCell nil} next:{NewCell nil} value:Value) end
proc {InsertAfter Node NewNode} Next = Node.next in (NewNode.next) := @Next (NewNode.prev) := Node case @Next of nil then skip [] node(prev:NextPrev ...) then NextPrev := NewNode end Next := NewNode end
A = {CreateNewNode a} B = {CreateNewNode b} C = {CreateNewNode c}
in
{InsertAfter A B} {InsertAfter A C} {Walk A Show} {WalkBackwards A Show}</lang>
Ruby
<lang ruby>class DListNode
def get_tail # parent class (ListNode) includes Enumerable, so the find method is available to us self.find {|node| node.succ.nil?} end
def each_previous(&b) yield self self.prev.each_previous(&b) if self.prev end
end
head = DListNode.from_array([:a, :b, :c]) head.each {|node| p node.value} head.get_tail.each_previous {|node| p node.value}</lang>
Tcl
Assuming that the List
class from this other task is already present...
<lang tcl># Modify the List class to add the iterator methods
oo::define List {
method foreach {varName script} { upvar 1 $varName v for {set node [self]} {$node ne ""} {set node [$node next]} { set v [$node value] uplevel 1 $script } } method revforeach {varName script} { upvar 1 $varName v for {set node [self]} {$node ne ""} {set node [$node previous]} { set v [$node value] uplevel 1 $script } }
}
- Demonstrating...
set first [List new a [List new b [List new c [set last [List new d]]]]] puts "Forward..." $first foreach char { puts $char } puts "Backward..." $last revforeach char { puts $char }</lang> Which produces this output:
Forward... a b c d Backward... d c b a
Python
This provides two solutions. One that explicitly builds a linked list and traverses it two ways, and another which uses pythons combined list/array class. Unless two lists explicitly needed to be spliced together in O(1) time, or a double linked list was needed for some other reason, most python programmers would probably use the second solution. <lang python>class List:
def __init__(self, data, next=None, prev=None): self.data = data self.next = next self.prev = prev
def append(self, data): if self.next == None: self.next = List(data, None, self) return self.next else: return self.next.append(data)
- Build the list
tail = head = List(10) for i in [ 20, 30, 40 ]:
tail = tail.append(i)
- Traverse forwards
node = head while node != None:
print(node.data) node = node.next
- Traverse Backwards
node = tail while node != None:
print(node.data) node = node.prev</lang>
This produces the desired output. However, I expect most python programmers would do the following instead:
<lang python>l = [ 10, 20, 30, 40 ] for i in l:
print(i)
for i in reversed(l): # reversed produces an iterator, so only O(1) memory is used
print(i)</lang>
Double-ended queues in python are provided by the collections.deque class and the array/list type can perform all the operations of a C++ vector (and more), so building one's own doubly-linked list would be restricted to very specialized situations.