LINKED LIST

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package il.ac.telhai.ds.stack;


public class DLinkedList<T> implements List<T> {
    DNode head;
    DNode cursor;
    DNode tail;

    public DLinkedList() {
        head = null;
        cursor = null;
        tail = null;
    }
    //insert function inserts a new element
    //the function checks if the list is empty and acts accordingly
    public void insert(T newElement) {
        if (newElement==null) throw new RuntimeException();
        DNode mynote = new DNode(newElement);
        //if is empty
        if (head == null)
        {
            head = mynote;
            head.next = null;
            head.prev = null;
            cursor = head;
            tail = head;

            return;
        }
        //if not empty
        mynote.setPrev(cursor);
        mynote.setNext(cursor.getNext());
        if (mynote.getNext()!=null) mynote.getNext().setPrev(mynote);
        cursor.setNext(mynote);
        if(mynote.getNext()==null) tail=mynote;
        cursor = mynote;
    }

    @Override
    //removes an item
    //when removing an item i distinguish beetwen 3 option:
    //first option : cursor points to tail and and tail is not head
    //second option : cursor points to head and its the only item
    //third option : cursor point to head and it is not the only item
    //fourth option: cursor points not to head and not to tail
    public T remove() {
        if (isEmpty())  return null;
        T element = cursor.getElement();
        //first
        if (cursor==tail && cursor!=head){
            cursor.getPrev().setNext(null);
            cursor = cursor.getPrev();
            tail = cursor;
            return element;
        }
        //if cursor is on head
        if (cursor==head )
        {
            //second
            if ( cursor.getNext()==null)
            {
                clear();
                return element;
            }
            //third
            cursor.getNext().setPrev(null);
            cursor =cursor.getNext();
            head = cursor;
            return element;
        }
        //fourth option
        cursor.getPrev().setNext(cursor.getNext());
        cursor.getNext().setPrev(cursor.getPrev());
        cursor = cursor.getNext();
        return element;




    }

Popularity 10/10 Helpfulness 1/10 Language java

Source: Grepper

Tags: java linked-list

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Contributed on Dec 30 2022

Mero

0 Answers Avg Quality 2/10

linked list

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// C++ program to convert a given Binary Tree to Doubly Linked List
#include <bits/stdc++.h>
 
// Structure for tree and linked list
struct Node {
    int data;
    Node *left, *right;
};
 
// Utility function for allocating node for Binary
// Tree.
Node* newNode(int data)
{
    Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return node;
}
 
// A simple recursive function to convert a given
// Binary tree to Doubly Linked List
// root    --> Root of Binary Tree
// head --> Pointer to head node of created doubly linked list
void BToDLL(Node* root, Node*& head)
{
    // Base cases
    if (root == NULL)
        return;
 
    // Recursively convert right subtree
    BToDLL(root->right, head);
 
    // insert root into DLL
    root->right = head;
 
    // Change left pointer of previous head
    if (head != NULL)
        head->left = root;
 
    // Change head of Doubly linked list
    head = root;
 
    // Recursively convert left subtree
    BToDLL(root->left, head);
}
 
// Utility function for printing double linked list.
void printList(Node* head)
{
    printf("Extracted Double Linked list is:\n");
    while (head) {
        printf("%d ", head->data);
        head = head->right;
    }
}
 
// Driver program to test above function
int main()
{
    /* Constructing below tree
            5
            / \
            3     6
        / \     \
        1 4     8
        / \     / \
        0 2     7 9 */
    Node* root = newNode(5);
    root->left = newNode(3);
    root->right = newNode(6);
    root->left->left = newNode(1);
    root->left->right = newNode(4);
    root->right->right = newNode(8);
    root->left->left->left = newNode(0);
    root->left->left->right = newNode(2);
    root->right->right->left = newNode(7);
    root->right->right->right = newNode(9);
 
    Node* head = NULL;
    BToDLL(root, head);
 
    printList(head);
 
    return 0;
}

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Source: www.geeksforgeeks.org

Tags: linked-list link

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Contributed on Apr 25 2022

Curious Caribou

0 Answers Avg Quality 2/10

linked list

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#include <iostream>
using namespace std;
 


class Node {
public:
    string name;
	string surName;
	int age;
	string gender;
    Node* next;
};
 
void Insert(Node** head, Node* newNode)
{
    Node* currentNode;
    
    if (*head == NULL|| (*head)->age >= newNode->age) {
        newNode->next = *head;
        *head = newNode;
    }
    else {
        
        currentNode = *head;
        while (currentNode->next != NULL && currentNode->next->age< newNode->age) {
            currentNode = currentNode->next;
        }
        newNode->next = currentNode->next;
        currentNode->next = newNode;
    }
}
 
Node* new_node(string name,string surName,	int age,	string gender)
{
    
    Node* newNode = new Node();
 
   
    newNode->name = name;
    newNode->surName = surName;
    newNode->age = age;
    newNode->gender = gender;
    newNode->next = NULL;
 
    return newNode;
}
 


void printList(Node* head)
{
    Node* temp = head;
    while (temp != NULL) {
        cout << temp->name << " "<<temp->surName<<"  "<<temp->age<<"  "<<temp->gender<<endl;
        temp = temp->next;
    }
}
 
 
void deleteNode(Node** head_ref, int key)
{
     
    
    Node* temp = *head_ref;
    Node* prev = NULL;
     
    
    if (temp != NULL && temp->age == key)
    {
        *head_ref = temp->next; 
        delete temp;            
        return;
    }
 
    
      else
    {
    while (temp != NULL && temp->age != key)
    {
        prev = temp;
        temp = temp->next;
    }
 
    
    if (temp == NULL)
        return;
 
   
    prev->next = temp->next;
 
    
    delete temp;
    }
}


int main()
{
    
    Node* head = NULL;
    Node* node = new_node("Donald", "Brown", 67, "M" );
    Insert(&head, node);
    node = new_node("Jason", "Travis", 90, "F");
    Insert(&head, node);
    node = new_node("Max", "Black", 27, "M");
    Insert(&head, node);
    node = new_node("Bobi", "Frank", 17, "F");
    Insert(&head, node);
   
    cout << "The list is sorted by gender in ascending order\n";
    printList(head);
    cout<<"After deleting a person who age is 17 the list becomes \n";
    deleteNode(&head, 17);
     printList(head);
     
     cout << "When a person whose is 20 is inserted the linked list becomes\n";
     node = new_node("Victoria", "Riberry", 20, "M");
    Insert(&head, node);
    printList(head);
    return 0;
}

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Source: www.assignmentexpert.com

Tags: linked-list link

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Contributed on Nov 13 2022

Winijnan Jitniyom

0 Answers Avg Quality 2/10

linked list

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#include<bits/stdc++.h>
using namespace std;

class node {
    public:
        int num;
        node* next;
        node(int val) {
            num = val;
            next = NULL;
        }
};

void insertNode(node* head,int val) {
    node* newNode = new node(val);
    if(head == NULL) {
        head = newNode;
        return;
    }
    
    node* temp = head;
    while(temp->next != NULL) temp = temp->next;
    
    temp->next = newNode;
    return;
}

node* reverse(node* ptr) {
    node* pre=NULL;
    node* nex=NULL;
    while(ptr!=NULL) {
        nex = ptr->next;
        ptr->next = pre;
        pre=ptr;
        ptr=nex;
    }
    return pre;
}

bool isPalindrome(node* head) {
    if(head==NULL||head->next==NULL) return true;
        
    node* slow = head;
    node* fast = head;
        
    while(fast->next!=NULL&&fast->next->next!=NULL) {
        slow = slow->next;
        fast = fast->next->next;
    }
        
    slow->next = reverse(slow->next);
    slow = slow->next;
    node* dummy = head;
        
    while(slow!=NULL) {
        if(dummy->num != slow->num) return false;
        dummy = dummy->next;
        slow = slow->next;
    }
    return true;
}

int main() {
    node* head = NULL;
    insertNode(head,1);
    insertNode(head,2);
    insertNode(head,3);
    insertNode(head,2);
    insertNode(head,1);
    isPalindrome(head)? cout<<"True" : cout<<"False";
    return 0;
}

Popularity 10/10 Helpfulness 1/10 Language cpp

Source: takeuforward.org

Tags: c++

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Contributed on Jul 24 2022

Precious Puma

0 Answers Avg Quality 2/10

linkedlist

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# A simple Python program to introduce a linked list
  
# Node class
class Node:
  
    # Function to initialise the node object
    def __init__(self, data):
        self.data = data # Assign data
        self.next = None # Initialize next as null
  
  
# Linked List class contains a Node object
class LinkedList:
  
    # Function to initialize head
    def __init__(self):
        self.head = None
  
  
# Code execution starts here
if __name__=='__main__':
  
    # Start with the empty list
    llist = LinkedList()
  
    llist.head = Node(1)
    second = Node(2)
    third = Node(3)
  
    '''
    Three nodes have been created.
    We have references to these three blocks as head,
    second and third
  
    llist.head     second             third
        |             |                 |
        |             |                 |
    +----+------+     +----+------+     +----+------+
    | 1 | None |     | 2 | None |     | 3 | None |
    +----+------+     +----+------+     +----+------+
    '''
  
    llist.head.next = second; # Link first node with second
  
    '''
    Now next of first Node refers to second. So they
    both are linked.
  
    llist.head     second             third
        |             |                 |
        |             |                 |
    +----+------+     +----+------+     +----+------+
    | 1 | o-------->| 2 | null |     | 3 | null |
    +----+------+     +----+------+     +----+------+
    '''
  
    second.next = third; # Link second node with the third node
  
    '''
    Now next of second Node refers to third. So all three
    nodes are linked.
  
    llist.head     second             third
        |             |                 |
        |             |                 |
    +----+------+     +----+------+     +----+------+
    | 1 | o-------->| 2 | o-------->| 3 | null |
    +----+------+     +----+------+     +----+------+
    '''

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Source: www.geeksforgeeks.org

Tags: linked-list link

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Contributed on Jun 17 2022

No Name

0 Answers Avg Quality 2/10

linked lists

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node_t * head = NULL;
head = (node_t *) malloc(sizeof(node_t));
if (head == NULL) {
    return 1;
}

head->val = 1;
head->next = NULL;

Popularity 10/10 Helpfulness 1/10 Language c

Source: www.learn-c.org

Tags: c

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Contributed on Jun 08 2021

Gentle Gharial

0 Answers Avg Quality 2/10

linked list

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package stack

// Node structure
type Node struct {
	Val  interface{}
	Next *Node
}

// Stack has jost top of node and with length
type Stack struct {
	top    *Node
	length int
}

// push add value to last index
func (ll *Stack) push(n interface{}) {
	newStack := &Node{} // new node

	newStack.Val = n
	newStack.Next = ll.top

	ll.top = newStack
	ll.length++
}

// pop remove last item as first output
func (ll *Stack) pop() interface{} {
	result := ll.top.Val
	if ll.top.Next == nil {
		ll.top = nil
	} else {
		ll.top.Val, ll.top.Next = ll.top.Next.Val, ll.top.Next.Next
	}

	ll.length--
	return result
}

// isEmpty to check our array is empty or not
func (ll *Stack) isEmpty() bool {
	return ll.length == 0
}

// len use to return length of our stack
func (ll *Stack) len() int {
	return ll.length
}

// peak return last input value
func (ll *Stack) peak() interface{} {
	return ll.top.Val
}

// show all value as an interface array
func (ll *Stack) show() (in []interface{}) {
	current := ll.top

	for current != nil {
		in = append(in, current.Val)
		current = current.Next
	}
	return
}

Popularity 10/10 Helpfulness 1/10 Language go

Source: github.com

Tags: go g

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Contributed on May 06 2022

gocrazygh

0 Answers Avg Quality 2/10

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