So the app reads from an external file a bunch of strings, each on a separate line.
For example:
and
cake
here
It is not arranged in any particular order. I need to read these letters and put them into linked list and finally sort them.
I need help on doing that:
Here is the current code:
import java.util.*;
import java.io.*;
public class LinkedList
{
static File dataInpt;
static Scanner inFile;
public static void main(String[] args) throws IOException
{
dataInpt=new File("C:\\lldata.txt");
inFile=new Scanner(dataInpt);
Node first = insertInOrder();
printList(first);
}
public static Node getNode(Object element)
{
Node temp=new Node();
temp.value=element;
temp.next=null;
return temp;
}
public static void printList(Node head)
{
Node ptr; //not pointing anywhere
for(ptr=head;ptr!=null;ptr=ptr.next)
System.out.println(ptr.value);
System.out.println();
}
public static Node insertInOrder()
{
Node first=getNode(inFile.next());
Node current=first,previous=null;
Node last=first;
int count=0;
while (inFile.hasNext())
{
if (previous!=null
&& ((String)current.value).compareTo((String)previous.value) > 0)
{
last.next=previous;
previous=last;
}
if (previous!=null
&& ((String)current.value).compareTo((String)previous.value) < 0)
{
current.next=last;
last=current;
}
previous=current;
current=getNode(inFile.next());
}
return last;
}
}
But that gives an infinite loop with "Cat".
Here is the data file:
Lol
Cake
Gel
Hi
Gee
Age
Rage
Tim
Where
And
Kite
Jam
Nickel
Cat
Ran
Jug
Here
Okay, self-study. Split the reading and inserting. Though old and new code both have 14 lines of code,
it makes it more intelligable.
public static Node insertInOrder() {
Node first = null;
while (inFile.hasNext()) {
String value = inFile.next().toString();
first = insert(first, value);
}
return first;
}
/**
* Insert in a sub-list, yielding a changed sub-list.
* #param node the sub-list.
* #param value
* #return the new sub-list (the head node might have been changed).
*/
private static Node insert(Node node, String value) {
if (node == null) { // End of list
return getNode(value);
}
int comparison = node.value.compareTo(value);
if (comparison >= 0) { // Or > 0 for stable sort.
Node newNode = getNode(value); // Insert in front.
newNode.next = node;
return newNode;
}
node.next = insert(node.next, value); // Insert in the rest.
return node;
}
This uses recursion (nested "rerunning"), calling insert inside insert. This works like a loop, or work delegation to a clone, or like a mathematical inductive proof.
Iterative alternative
also simplified a bit.
private static void Node insert(Node list, String value) {
Node node = list;
Node previous = null;
for (;;) {
if (node == null || node.value.compareTo(value) >= 0) {
Node newNode = getNode(value);
newNode.next = node;
if (previous == null)
list = newNode;
else
previous.next = newNode;
break;
}
// Insert in the rest:
previous = node;
node = node.next;
}
return list;
}
public static Node insertInOrder()
{
Node first=getNode(inFile.next());
Node current=first,previous=null;
Node last=first;
int count=0;
while (inFile.hasNext())
{
if (previous!=null
&& ((String)current.value).compareTo((String)previous.value) > 0)
{
last.next=previous;
previous=last;
}
if (previous!=null
&& ((String)current.value).compareTo((String)previous.value) < 0)
{
current.next=last;
last=current;
}
previous=current;
current=getNode(inFile.next());
}
return last;
}
First of all, you never do anything with the last line read from the file, so that's not ever inserted. You have to read the line and create the new Node before relinking next pointers.
Then, if last and previous refer to the same Node and the data of current is larger than that of previous,
if (previous!=null
&& ((String)current.value).compareTo((String)previous.value) > 0)
{
last.next=previous;
previous=last;
}
You set last.next = last, breaking the list. From the code (in particular the absence of a sort(Node) function), it seems as though you want to sort the list as it is created. But you only ever compare each new Node with one other, so that doesn't maintain order.
For each new node, you have to find the node after which it has to be inserted, scanning from the front of the list, and modify current.next and the predecessor's next.
In relatively simple code like that in your question, a good exercise to understanding it is to work through a few interations of your loop, inspecting the values of all your local variable to see the effect of your code. You can even do it by hand if the code is simple. If it is too difficult to do by hand, your code is probably too complicated. If you can't follow it, how can you know if you are doing what you intend. For example, I could be wrong, but this appears the be the state at the top of each iteration of the loop. It starts falling apart on the third time through, and by the fourth you have a severe problem as your list becomes disjointed.
1)last = first = Lol, current = previous = null
Lol->null
2)last = first = previous = Lol, current = Cake
Lol->Lol
3)first = Lol, last = Cake, previous = Cake, current = Gel
Cake->Lol->Lol
4)first = Lol, last = Cake, previous = Cake, current = Hi
Cake->Gel, Lol->Lol
Quite honestly, if I were running the course, I would consider the correct answer to be:
List<String> list = new LinkedList<String>();
// read in lines and: list.add(word);
Collections.sort(list);
Ok, I don't remember exactly school theory about insertion sort, but here is somehow a mix of what I think it is and your code:
import java.io.File;
import java.io.IOException;
import java.util.Scanner;
public class LinkedList {
public static class Node {
public String value;
public Node next;
}
static File dataInpt;
static Scanner inFile;
public static void main(String[] args) throws IOException {
inFile = new Scanner("Lol\r\n" + "Cake\r\n" + "Gel\r\n" + "Hi\r\n" + "Gee\r\n" + "Age\r\n" + "Rage\r\n" + "Tim\r\n" + "Where\r\n"
+ "And\r\n" + "Kite\r\n" + "Jam\r\n" + "Nickel\r\n" + "Cat\r\n" + "Ran\r\n" + "Jug\r\n" + "Here");
Node first = insertInOrder();
printList(first);
}
public static Node getNode(String element) {
Node temp = new Node();
temp.value = element;
temp.next = null;
return temp;
}
public static void printList(Node head) {
Node ptr; // not pointing anywhere
for (ptr = head; ptr != null; ptr = ptr.next) {
System.out.println(ptr.value);
}
System.out.println();
}
public static Node insertInOrder() {
Node current = getNode(inFile.next());
Node first = current, last = current;
while (inFile.hasNext()) {
if (first != null && current.value.compareTo(first.value) < 0) {
current.next = first;
first = current;
} else if (last != null && current.value.compareTo(last.value) > 0) {
last.next = current;
last = current;
} else {
Node temp = first;
while (current.value.compareTo(temp.value) < 0) {
temp = temp.next;
}
current.next = temp.next;
temp.next = current;
}
current = getNode(inFile.next());
}
return first;
}
}
And it works like a charm. Of course this far from optimal, both in terms of performance and code reuse.
Related
I am trying to write a program that takes an integer and inserts it into the second position in the linked list, but when I run my program it prints out (127, ) when I want it to print out (12, 7). I know it is probably an easy fix, but I am unsure of how to fix this small error. I have tried switching words around like head and tail and next, but nothing I do seems to work. Any input would help a ton!
public class IntSinglyLinkedList {
private static class Node {
private Integer element;
private Node next;
public Node(Integer e, Node n) {
element = e;
next = n;
}
private Node head = null;
private Node tail = null;
private int size = 0;
public IntSinglyLinkedList() { }
public int size() { return size; }
public void addSecond(Integer e) {
if(head == null)
return;
Node Final = new Node(e, head);
Final.next = head.next;
head.next = Final;
}
public static void main(String[] args) {
IntSinglyLinkedList sl = new IntSinglyLinkedList();
sl.addFirst(12);
sl.addSecond(7);
System.out.println(sl.toString());
}
}
My toString also looks like this if this is any help!
public String toString() {
StringBuilder sb = new StringBuilder("(");
Node walk = head;
while (walk != null) {
sb.append(walk.getElement());
if (walk != tail)
sb.append(", ");
walk = walk.getNext();
}
sb.append(")");
return sb.toString();
}
my addFirst method
public void addFirst(Integer e) {
head = new Node(e, head);
if (size == 0)
tail = head;
size++;
}
Why aren't you just using a LinkedList?
LinkedList<Integer> ll = new LinkedList<>();
ll.add(12);
ll.add(1, 7);
System.out.println("LinkedList is: " + ll.toString());
Output would be: LinkedList is: [12, 7]
You never update your tail element after your addFirst, so the , will not be printed after the first element.
You have to add tail = Final; in your addSecond method to make this particular example to print correctly.
That answers the specific question of why you get (127,) printed instead of (12, 7) in this particular case.
However, it is hard to see where you are going with this. Are you planning to write a new method for every element you want to add? It would make more sense to have a general add:
public Node add(final Integer i) {
final var node = new Node(i, null);
if (size == 0) {
head = node;
} else {
tail.next = node;
}
tail = node;
size++;
return node;
}
I'm newbie in programming and I'm practicing a Java Programming Language. I was having a rough day in finding the solution of my program because I cannot get my "next" pointer and I really want to print my last value. Could someone help me to fix this and explain to me? Thank you in advance. Here's my code.
Note: The output of my program is 5.
public class Node {
private int data;
private Node next;
public Node (int data){
this.data = data;
}
public int getData() {
return this.data;
}
public void setNext(Node n) {
this.next = n;
}
public Node getNext() {
return this.next;
}
}
public class LinkedList {
private static Node head, next;
public LinkedList (int data) {
head = new Node (data);
}
public void addLast(int data) {
Node n = new Node (data);
if (head == null) {
head = n;
}
else {
Node temp = head;
temp.setNext(next);
while (temp.getNext() != null) {
temp = temp.getNext();
}
Node t = temp.getNext();
t = n;
}
}
public void printList() {
head.setNext(next);
while (head.getNext() != null) {
System.out.println(head.getData());
head = head.getNext();
}
System.out.println(head.getData());
}
public static void main(String[] args) {
LinkedList l = new LinkedList(5);
l.addLast(7);
l.printList();
}
}
I suggest following two amendment to your code.
with following else block in addLast method.
Node temp = head;
temp.setNext(next); // this line causing the next object to be set to null all the time. commenting this line will help in making sure the follwing loop reaches to end of the list, otherwise the while loop will always exit without any iteration.
while (temp.getNext() != null) {
temp = temp.getNext();
}
Node t = temp.getNext();
t = n; // this will also not change the linking. Its basically assigned a new value to t.
use following suggestion
Node temp = head;
while (temp.getNext() != null) {
temp = temp.getNext();
}
// now we reached end of list and temp.next is null.
// assign newly createdd node to temp.next
temp.setNext(n);
While iterating the element in printList same problem exist as mentioned in point 1. try to use following suggestion for printList method.
// head.setNext(next); // This line will always set head.next to null and whole list will be lost. Instead of this use following line
Node temp = head;
while (temp.getNext() != null) { // here if you use head its position will move to end. So use temp variable for iteration
System.out.println(temp.getData());
temp= temp.getNext();
}
System.out.println(temp.getData());
You may also need to study list iteration algorithm to have better understanding.
I've made some amendments to make your code work. The If statement in your addLast method:
if (head == null) {
is redundant since your LinkedList can only be initialized by passing some data, hence head will never be null, it will always point to the Node containing data
Also the line
head.setNext(next);
in your printList() was problematic, it was always pointing to null
public class LinkedList {
private static Node head, next;
public LinkedList(int data) {
head = new Node(data);
}
public void addLast(int data) {
Node n = new Node(data);
Node temp = head;
temp.setNext(next);
while (temp.getNext() != null) {
temp = temp.getNext();
}
temp.setNext(n);
}
public void printList() {
while (head.getNext() != null) {
System.out.println(head.getData());
head = head.getNext();
}
System.out.println(head.getData());
}
public static void main(String[] args) {
LinkedList l = new LinkedList(5);
l.addLast(7);
l.printList();
}
}
TL;DR:
You set null as the next node in your printList() method;
Your addLast does not work either (you do not set the next node (see details below);
You should never set the node (or do any logical alteration whatsoever) in your print method. Print must just print, as the name suggests, and it should not contain any side-effect, amending your data structure. That is: you have to clearly separate your concerns.
In your current addLast, you do:
public void addLast(int data) {
Node n = new Node (data);
if (head == null) {
head = n;
}
else {
Node temp = head;
temp.setNext(next);
while (temp.getNext() != null) {
temp = temp.getNext();
}
Node t = temp.getNext();
t = n;
}
}
which means, that when your temp's next node is null, you never add the Node you instantiate with your int argument.
Change the else block as follows:
else {
Node temp = head;
temp.setNext(next);
while (temp.getNext() != null) {
temp = temp.getNext();
}
temp.setNext(n);
//two redundant lines removed
}
Correspondingly, remove head.setNext(next); (and possibly unnecessary System.out.println() statement) from your printList() method.
P. S. I would really recommend you to spend some time on the Linked List Data Structure (Data Structure, and not the Java code), as your current design, shows that you need to have a better grasp of it.
Given singly Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> null
Modify middle element as doubly Linked List Node
here middle element is 3
3 -> next should point to 4
3 -> prev should point to 1
Can any one suggest how can it be done ? interviewer gave me hint use interface. but I couldn't figure it out how.
I have to iterate over this linked list and print all the node and when it reaches to the middle, print where next and prev is pointing to, then print till the end of the list.
Expected output : 1, 2, Middle: 3, Prev: 1, Next: 4, 5
I'm facing problem in adding the middle node.
So, this "works", but if this is expected to be answered on an interview, it is way too much work.
LinkedList
public class LinkedList {
public interface Linkable<V, L extends Linkable> {
V getValue();
L getNext();
void setNext(L link);
}
public static class Node implements Linkable<Integer, Linkable> {
int value;
Linkable next;
Node(int value) {
this.value = value;
}
#Override
public Integer getValue() {
return value;
}
#Override
public Linkable getNext() {
return next;
}
#Override
public void setNext(Linkable link) {
this.next = link;
}
}
private Linkable head;
public boolean isEmpty() {
return this.head == null;
}
public Linkable getHead() {
return head;
}
public void add(int v) {
Node next = new Node(v);
if (isEmpty()) {
this.head = next;
} else {
Linkable tmp = this.head;
while (tmp.getNext() != null) {
tmp = tmp.getNext();
}
tmp.setNext(next);
}
}
}
Interface
interface DoublyLinkable<V, L extends LinkedList.Linkable> extends LinkedList.Linkable<V,L> {
LinkedList.Linkable getPrev();
void setPrev(LinkedList.Linkable prev);
}
DoubleNode
public class DoubleNode extends LinkedList.Node implements DoublyLinkable<Integer, LinkedList.Linkable> {
LinkedList.Linkable prev;
public DoubleNode(int value) {
super(value);
}
#Override
public LinkedList.Linkable getPrev() {
return prev;
}
#Override
public void setPrev(LinkedList.Linkable prev) {
this.prev = prev;
}
}
Driver
Outputs
1, 2, Middle: 3, Prev: 1, Next: 4, 5
public class Driver {
public static LinkedList getList() {
LinkedList list = new LinkedList();
for (int i = 1; i <= 5; i++) {
list.add(i);
}
return list;
}
public static void main(String[] args) {
LinkedList list = getList();
LinkedList.Linkable head = list.getHead();
LinkedList.Linkable beforeMiddle = null;
LinkedList.Linkable middle = list.getHead();
LinkedList.Linkable end = list.getHead();
if (head != null) {
// find the middle of the list
while (true) {
if (end.getNext() == null || end.getNext().getNext() == null) break;
beforeMiddle = middle;
middle = middle.getNext();
end = end.getNext().getNext();
}
// Replace middle by reassigning the pointer to it
if (beforeMiddle != null) {
DoubleNode n = new DoubleNode((int) middle.getValue()); // same value
n.setPrev(list.getHead()); // point back to the front
n.setNext(middle.getNext()); // point forward to original value
beforeMiddle.setNext((DoublyLinkable) n);
middle = beforeMiddle.getNext();
}
// Build the "expected" output
StringBuilder sb = new StringBuilder();
final String DELIMITER = ", ";
head = list.getHead();
boolean atMiddle = false;
if (head != null) {
do {
if (head instanceof DoublyLinkable) {
atMiddle = true;
String out = String.format("Middle: %d, Prev: %d, ", (int) head.getValue(), (int) ((DoublyLinkable) head).getPrev().getValue());
sb.append(out);
} else {
if (atMiddle) {
sb.append("Next: ");
atMiddle = false;
}
sb.append(head.getValue()).append(DELIMITER);
}
head = head.getNext();
} while (head != null);
}
sb.setLength(sb.length() - DELIMITER.length());
System.out.println(sb.toString());
}
}
}
By definition, a single-linked list consists of single-linked nodes only, and a double-linked consists of double-linked nodes only. Otherwise. it is neither.
By definition the field prev of a double-linked list must point to the previous element.
Whatever you are supposed to build. It's something not well specified. So if you really were asked this in an interview (and did not misunderstand the question - maybe he wanted you to point out that ghis violates the interface?) this is a case for the code horror stories of http://thedailywtf.com/ - section "incompetent interviewers".
If you haven't, you'd better define a lenght() function so given one linked list you can know how many nodes does it have.
Thanks to the response of Cereal_Killer to the previous version of this answer, I noticed that the list is firstly a singly linked list, and you just have to make the middle node be linked both to the next node and to some previous node.
Now I guess that you have defined two structures (Struct, Class or whatever depending on the language you're using). So lets say you have Node_s defined as a node with only a next pointer, and Node_d with both a next and a prev pointer. (Node_d may inherite from Node_s so you just have to add the prev attribute in the child class). Knowing this, the code above should be doing what you need:
function do_it(my_LinkedList linkedList){
int i_middle;
int length = linkedList.length();
if ( (length รท 2 ) != 0 ) i_middle = length / 2;
else return -1;
Node_s aux = linkedList.first();
int index = 0;
Node_d middle= null;
while (aux != null) {
if (index == i_middle - 1){ //now aux is the middle's previous node
middle.data = aux.next.data; //aux.next is the middle singly node, we assignate the data to the new double linked node
middle.prev = aux; //as we said, aux is the prev to the middle
midle.next = aux.next.next; //so aux.next.next is the next to the middle
print(what you need to print);
}else {
print("Node " + index " next: "+ aux.next);
}//end if
index++;
aux = aux.next;
} //end while
}//end function
This previous code should be doing what you need. I wrote the answer in some kind of pseudo-java code so if you're not familiar with Java or don't understand what my pseudo-code does, please let me know. Anyway, the idea of my code may present some troubles depending on the language you're working with, so you'll have to adapt it.
Note that at the end of the execution of this program, your data structure won't be a singly linked list, and neither a double one, since you'll have linkedList.length() - 1 nodes linked in a signly way but the middle one will have two links.
Hope this helps.
I've been working through some standard coding interview questions from a book I recently bought, and I came across the following question and answer:
Implement an algorithm to find the nth to last element in a linked list.
Here's the provided answer:
public static LinkedListNode findNtoLast(LinkedListNode head, int n) { //changing LinkedListNode to ListNode<String>
if(head == null || n < 1) {
return null;
}
LinkedListNode p1 = head;
LinkedListNode p2 = head;
for(int j = 0; j < n-1; ++j) {
if(p2 == null) {
return null;
}
p2 = p2.next;
}
if(p2 == null) {
return null;
}
while(p2.next != null) {
p1 = p1.next;
p2 = p2.next;
}
return p1;
}
I understand the algorithm, how it works, and why the book lists this as its answer, but I'm confused about how to access the LinkedListNodes to send as an argument to the method. I know that I'd have to create a LinkedListNode class (since Java doesn't already have one), but I can't seem to figure out how to do that. It's frustrating because I feel like I should know how to do this. Here's something that I've been working on. I'd greatly appreciate any clarification. You can expand/comment on my code or offer your own alternatives. Thanks.
class ListNode<E> {
ListNode<E> next;
E data;
public ListNode(E value) {
data = value;
next = null;
}
public ListNode(E value, ListNode<E> n) {
data = value;
next = n;
}
public void setNext(ListNode<E> n) {
next = n;
}
}
public class MyLinkedList<E> extends LinkedList {
LinkedList<ListNode<E>> list;
ListNode<E> head;
ListNode<E> tail;
ListNode<E> current;
ListNode<E> prev;
public MyLinkedList() {
list = null;
head = null;
tail = null;
current = null;
prev = null;
}
public MyLinkedList(LinkedList<E> paramList) {
list = (LinkedList<ListNode<E>>) paramList; //or maybe create a loop assigning each ListNode a value and next ptr
head = list.getFirst();
tail = list.getLast(); //will need to update tail every time add new node
current = null;
prev = null;
}
public void addNode(E value) {
super.add(value);
//ListNode<E> temp = tail;
current = new ListNode<E>(value);
tail.setNext(current);
tail = current;
}
public LinkedList<ListNode<E>> getList() {
return list;
}
public ListNode<E> getHead() {
return head;
}
public ListNode<E> getTail() {
return tail;
}
public ListNode<E> getCurrent() {
return current;
}
public ListNode<E> getPrev() {
return prev;
}
}
How can the LinkedListNode head from a LinkedList?
Update: I think part of my confusion comes from what to put in the main method. Do I need to create a LinkedList of ListNode? If I do that, how would I connect the ListNodes to each other? How would I connect them without using a LinkedList collection object? If someone could show me how they would code the main method, I think that would put things into enough perspective for me to solve my issues. Here's my latest attempt at the main method:
public static void main(String args[]) {
LinkedList<ListNode<String>> list = new LinkedList<ListNode<String>>();
//MyLinkedList<ListNode<String>> list = new MyLinkedList(linkedList);
list.add(new ListNode<String>("Jeff"));
list.add(new ListNode<String>("Brian"));
list.add(new ListNode<String>("Negin"));
list.add(new ListNode<String>("Alex"));
list.add(new ListNode<String>("Alaina"));
int n = 3;
//ListIterator<String> itr1 = list.listIterator();
//ListIterator<String> itr2 = list.listIterator();
LinkedListNode<String> head = new LinkedListNode(list.getFirst(), null);
//String result = findNtoLast(itr1, itr2, n);
//System.out.println("The " + n + "th to the last value: " + result);
//LinkedListNode<String> nth = findNtoLast(list.getFirst(), n);
ListNode<String> nth = findNtoLast(list.getFirst(), n);
System.out.println("The " + n + "th to the last value: " + nth);
}
In an attempt to connect the nodes without using a custom linked list class, I have edited my ListNode class to the following:
class ListNode<E> {
ListNode<E> next;
ListNode<E> prev; //only used for linking nodes in singly linked list
ListNode<E> current; //also only used for linking nodes in singly linked list
E data;
private static int size = 0;
public ListNode() {
data = null;
next = null;
current = null;
if(size > 0) { //changed from prev != null because no code to make prev not null
prev.setNext(this);
}
size++;
}
public ListNode(E value) {
data = value;
next = null;
current = this;
System.out.println("current is " + current);
if(size > 0) {
prev.setNext(current);//this line causing npe
}
else
{
prev = current;
System.out.println("prev now set to " + prev);
}
size++;
System.out.println("after constructor, size is " + size);
}
public ListNode(E value, ListNode<E> n) {
data = value;
next = n;
current = this;
if(size > 0) {
prev.setNext(this);
}
size++;
}
public void setNext(ListNode<E> n) {
next = n;
}
}
As is right now, the program will run until it reaches prev.setNext(current); in the single argument constructor for ListNode. Neither current nor prev are null at the time this line is reached. Any advice would be greatly appreciated. Thanks.
You don't actually need a separate LinkedList class; the ListNode class is a linked list. Or, to state it differently, a reference to the head of the list is a reference to the list.
The use of head, tail, current, prev in the sample code you posted has come from a double-linked list which is a data type that has links in both directions. This is more efficient for certain types of applications (such as finding the nth last item).
So I would recommend renaming your ListNode class to LinkedList and renaming next to tail.
To add a new item to the list you need a method that creates a new list with the new item at it's head. Here is an example:
class LinkedList<E> {
...
private LinkedList(E value, LinkedList<E> tail) {
this.data = value;
this.tail = tail;
}
public LinkedList<E> prependItem(E item) {
return new LinkedList(item, this);
}
}
Then to add a new item i to list you use list = list.prependItem(i);
If for some reason you need to always add the items to the end, then:
private LinkedList(E value) {
this.data = value;
this.tail = null;
}
public void appendItem(E item) {
LinkedList<E> list = this;
while (list.tail != null)
list = list.tail;
list.tail = new LinkedList<>(item);
}
However this is obviously pretty inefficient for long lists. If you need to do this then either use a different data structure or just reverse the list when you have finished adding to it.
Incidentally, an interesting side effect of this is that a reference to any item in the list is a reference to a linked list. This makes recursion very easy. For example, here's a recursive solution for finding the length of a list:
public int getLength(LinkedList list) {
if (list == null) {
return 0;
} else {
return 1 + getLength(list.getTail());
}
}
And using this a simple (but very inefficient!) solution to the problem you provided - I've renamed the method to make its function more obvious:
public LinkedList getTailOfListOfLengthN(LinkedList list, int n) {
int length = getLength(list);
if (length < n) {
return null;
} else if (length == n) {
return list;
} else {
return getTailOfLengthN(list.getTail(), n);
}
}
And to reverse the list:
public LinkedList<E> reverse() {
if (tail == null) {
return this;
} else {
LinkedList<E> list = reverse(tail);
tail.tail = this;
tail = null;
return list;
}
}
As I hope you can see this makes the methods a lot more elegant than separating the node list classes.
Actually you have created a linked list with you class ListNode.
A linked list is made of a node and a reference to another linked list (see the recursion?).
My assigment is to create my own linked list class (I can't use Java's LinkedList class) and implement a selection sort on it by swapping pointers rather than data.
I've created a double-linked MyLinkedList class, but I'm having trouble with the sort method. I've tried a number of things but nothing has worked - not even anything that would make sense to post here for correction. (I do know that I need to use at least one temp Node.) It has to be a selection sort.
I'm not looking for someone to code it for me, necessarily; I'm hoping someone can help me with an algorithm that I can then turn into code myself. Any help is appreciated greatly.
Here's how I've implemented the MyLinkedList class and the associated Node class:
public class MyLinkedList
{
private Node head;
private int count;
public MyLinkedList()
{
head = new Node(null);
count = 0;
}
public void add(String line)
{
Node temp = new Node(line);
Node current = head;
while (current.getNext() != null)
{
current = current.getNext();
}
temp.setLine (line); // not sure this is how to do it
current.setNext(temp);
temp.setPrev(current);
count++;
}
public void displayList()
{
Node current = head;
for (int i = 0; i < count; i++)
{
current = current.getNext();
System.out.println(current.getLine());
}
}
public void sortList()
{
Node start = head;
Node index = start;
Node min = start;
Node temp1, temp2;
while (start.getNext() != null)
{
index = index.getNext();
if (index.getLine().compareTo(min.getLine()) < 0)
{
min = index;
}
//swap - HELP, PLEASE :-)
{
// Algorithm???
}
}
}
public int size()
{
return count;
}
private class Node
{
String textLine;
Node next;
Node prev;
public Node()
{
textLine = null;
next = null;
prev = null;
}
public Node (String line)
{
textLine = (line);
next = null;
prev = null;
}
public Node (String line, Node node1, Node node2)
{
textLine = line;
prev = node1;
next = node2;
}
public String getLine()
{
return textLine;
}
public Node getNext()
{
return next;
}
public Node getPrev()
{
return prev;
}
public void setLine(String line)
{
textLine = line;
}
public void setNext(Node nextNode)
{
next = nextNode;
}
public void setPrev(Node prevNode)
{
prev = prevNode;
}
}
}
It may get confusing if an empty MyLinked List has a node in it even if it's just one with null prev, next and data, so you need to be careful of that MyLinkedList constructor - it would probably be much easier if it read simply head = null;.
Also it would be useful if a MyLinked List had a tail node as well to save you following the chain to the end to find where add should put a new Node.
After that, I think the problem is that you haven't noticed you need two loops: one to work your way through the list to keep track of where the unsorted nodes start, and one to find the smallest node from thereon. You also need to write a swap method for Node so that you can write something like this untested pseudocode that just happens to look a lot like Java
for (index = head; index != null; index = index.getNext()) {
min = index;
for (test = min.getNext(); test != null; test = test.getNext) {
if (test.getLine().compareTo(min.getLine()) < 0)
min = test;
}
if (min != index) {
swap(index, min);
index = min;
}
}
and swap would look roughly like
public void swap(Node other)
{
Node temp;
temp = next;
next = other.getNext();
other.setNext(temp);
temp = prev;
prev = other.getPrev();
other.setPrev(temp);
other.getNext().setPrev(this);
other.getPrev().setNext(this);
this.getNext().setPrev(other);
this.getPrev().setNext(other);
}
Note again this is completely untested and hasn't even seen a compiler.
Make sure to think about special cases like when the list is empty or has only one element in it, and when there is only one node left unsorted in the list.
I couldn't leave this without pointing out that swap is actually a lot more complex than that. I've added a few lines to correct the pointers in the nodes before and after the nodes to be swapped. You also need to consider:
Whether either of the nodes that are swapped are at the end of the list, in which case the head (and tail if you have one) of the list will need to be updated instead of the pointers in the adjacent nodes. That's fairly obvious.
Whether the nodes to be swapped are next to each other in the list, when if you apply the normal algorithm you get nodes pointing to themselves. That's less obvious.