code ----- public void add(int key, T value) { Link previou.pdf
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code: ----- public void add(int key, T value) { Link previous = null; // start at first Link current = first; // until end of list, while( current != null && key > current.getKey() ) { // or current > key, previous = current; current = current.next; // go to next item } if(previous==null) // if beginning of list, first = theLink; // first --> new link else // not at beginning, previous.next = theLink; // prev --> new link theLink.next = current; // new link --> current } Pseudo code: ------------ 1. start from first node Link current=first; 2. Traverse the hash table till the end of the list and the key value becomes greater than the current key value, and locate a position,called previous. 3. If the obtained position is null, make the new node as first node. Place the value in that node. 4. if the position is not null,insert the new node as next to previous node. Solution code: ----- public void add(int key, T value) { Link previous = null; // start at first Link current = first; // until end of list, while( current != null && key > current.getKey() ) { // or current > key, previous = current; current = current.next; // go to next item } if(previous==null) // if beginning of list, first = theLink; // first --> new link else // not at beginning, previous.next = theLink; // prev --> new link theLink.next = current; // new link --> current } Pseudo code: ------------ 1. start from first node Link current=first; 2. Traverse the hash table till the end of the list and the key value becomes greater than the current key value, and locate a position,called previous. 3. If the obtained position is null, make the new node as first node. Place the value in that node. 4. if the position is not null,insert the new node as next to previous node..
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This assignment and the next (#5) involve design and development of a.pdf
This assignment and the next (#5) involve design and development of a sequential
non contiguous and dynamic datastructure called LinkedList. A linked list object is
a container consisting of connected ListNode objects. As before, we are not going
to use pre-fabricated classes from the c++ library, but construct the LinkedList
ADT from scratch.
The first step is construction and testing of the ListNode class. A ListNode object
contains a data field and a pointer to the next ListNode object (note the recursive
definition).
#This assignment requires you to
1. Read the Assignment 4 Notes
2. Watch the Assignment 4 Support video
3. Implement the following methods of the ListNode class
-custom constructor
-setters for next pointer and data
4. Implement the insert and remove method in the main program
5. Scan the given template to find the above //TODO and implement the code
needed
//TODO in ListNodecpp.h file
public: ListNode(T idata, ListNode<T> * newNext);
public: void setNext(ListNode<T> * newNext);
public: void setData(T newData);
// TODO in main program
void remove(ListNode<int> * &front,int value)
void insert(ListNode<int> * &front,int value)
# The driver is given ListNodeMain.cpp is given to you that does the following
tests
1. Declares a pointer called front to point to a ListNode of datatype integer
2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked
list.
3. Inserts ListNode with data 3 to the list
4. Removes node 1 and adds it back to test removing and adding the first
element
5. Removes node 3 to test removing a middle node
6. Removes node 4 to test removing the last node
7. Attempt to remove a non existent node
8. Remove all existing nodes to empty the list
9. Insert node 4 and then node 1 to test if insertions preserve order
10.Print the list
Main.cpp
#include <iostream>
#include "ListNodecpp.h"
// REMEMBER each ListNode has two parts : a data field
// and an address field. The address is either null or points to the next node
//in the chain
//Requires: integer value for searching, address of front
//Effects: traverses the list node chain starting from front until the end comparing search value
with listnode getData. Returns the original search value if found, if not adds +1 to indicate not
found
//Modifies: Nothing
int search(ListNode<int> * front, int value);
//Requires: integer value for inserting, address of front
//Effects: creates a new ListNode with value and inserts in proper position (increasing order)in
the chain. If chain is empty, adds to the beginning
//Modifies: front, if node is added at the beginning.
//Also changes the next pointer of the previous node to point to the
//newly inserted list node. the next pointer of the newly inserted pointer
//points to what was the next of the previous node.
//This way both previous and current links are adjusted
//******** NOTE the use of & in passing pointer to front as parameter -
// Why do you think this is needed ?**********
void insert(ListNode<int> * &fr.
please i need help Im writing a program to test the merge sort alg.pdf
please i need help I\'m writing a program to test the merge sort algorithm for linked lists.
I need to Print the list before being sorted and the list after sorting the elements
here is my code only help in test program the four classes are ok except the test program.
TEST PROGRAM
import java.io.*;
import java.util.*;
public class Ch9_ProgExercise7 {
Scanner input = new Scanner(System.in);
public static void main(String[] args) throws IOException {
OrderedLinkedList list
= new OrderedLinkedList(); //Line 1
IntElement num = new IntElement(); //Line 2
// Prompt the user to enter some integers
System.out.println(\"Enter integers numbers \");
String temp = input.hasnextLine();
String[] split = temp.split(\" \");
for (int i = 0; i < split.length; i++) {
Integer a = Integer.parseInt(split[i]);
// insert the intgers into the list you created above
num.setNum(a);
list.insertNode(num);
}
System.out.println();
// Print the list before being sorted
// call the mergesort method
System.out.println();
// print the list after sorting the elements
}
}
ORDEREDLICKEDLIST CLASS
public class OrderedLinkedList extends LinkedListClass
{
//default constructor
public OrderedLinkedList()
{
super();
}
//copy constructor
public OrderedLinkedList(OrderedLinkedList otherList)
{
super(otherList);
}
public void linkedInsertionSort()
{
LinkedListNode lastInOrder;
LinkedListNode firstOutOfOrder;
LinkedListNode current;
LinkedListNode trailCurrent;
lastInOrder = first;
if(first == null)
System.out.println(\"Cannot sort an empty list\");
else
if(first.link == null)
System.out.println(\"The list is of length 1. \"
+ \"Already in order\");
else
while(lastInOrder.link != null)
{
firstOutOfOrder = lastInOrder.link;
if(firstOutOfOrder.info.compareTo(first.info) < 0)
{
lastInOrder.link = firstOutOfOrder.link;
firstOutOfOrder.link = first;
first = firstOutOfOrder;
}
else
{
trailCurrent = first;
current = first.link;
while(current.info.compareTo(firstOutOfOrder.info) < 0)
{
trailCurrent = current;
current = current.link;
}
if(current != firstOutOfOrder)
{
lastInOrder.link = firstOutOfOrder.link;
firstOutOfOrder.link = current;
trailCurrent.link = firstOutOfOrder;
}
else
lastInOrder = lastInOrder.link;
}
}
}//end linkedInsertionSort
//Method to determine whether searchItem is in
//the list.
//Postcondition: Returns true if searchItem is found
// in the list; otherwise, it returns
// false.
public boolean search(DataElement searchItem)
{
LinkedListNode current; //variable to traverse the list
boolean found;
current = first; //set current to point to the first
//node in the list
found = false; //set found to false
while(current != null && !found ) //search the list
if(current.info.compareTo(searchItem) >= 0)
found = true;
else
current = current.link; //make current point to
//the next node
if(found)
found = current.info.equals(searchItem);
return found;
}
//Method to insert insertItem in the list
//Postcondition: first points to the new list,
// newItem is inserted at the proper place
//.
The LinkedList1 class implements a Linked list. class.pdf
/**
The LinkedList1 class implements a Linked list.
*/
class LinkedList1
{
/**
The Node class stores a list element
and a reference to the next node.
*/
private class Node
{
String value;
Node next;
/**
Constructor.
@param val The element to store in the node.
@param n The reference to the successor node.
*/
Node(String val, Node n)
{
value = val;
next = n;
}
/**
Constructor.
@param val The element to store in the node.
*/
Node(String val)
{
// Call the other (sister) constructor.
this(val, null);
}
}
private Node first; // list head
private Node last; // last element in list
/**
Constructor.
*/
public LinkedList1()
{
first = null;
last = null;
}
/**
The isEmpty method checks to see
if the list is empty.
@return true if list is empty,
false otherwise.
*/
public boolean isEmpty()
{
return first == null;
}
/**
The size method returns the length of the list.
@return The number of elements in the list.
*/
public int size()
{
int count = 0;
Node p = first;
while (p != null)
{
// There is an element at p
count ++;
p = p.next;
}
return count;
}
/**
The add method adds an element to
the end of the list.
@param e The value to add to the
end of the list.
*/
public void add(String e)
{
if (isEmpty())
{
first = new Node(e);
last = first;
}
else
{
// Add to end of existing list
last.next = new Node(e);
last = last.next;
}
}
/**
The add method adds an element at a position.
@param e The element to add to the list.
@param index The position at which to add
the element.
@exception IndexOutOfBoundsException When
index is out of bounds.
*/
public void add(int index, String e)
{
if (index < 0 || index > size())
{
String message = String.valueOf(index);
throw new IndexOutOfBoundsException(message);
}
// Index is at least 0
if (index == 0)
{
// New element goes at beginning
first = new Node(e, first);
if (last == null)
last = first;
return;
}
// Set a reference pred to point to the node that
// will be the predecessor of the new node
Node pred = first;
for (int k = 1; k <= index - 1; k++)
{
pred = pred.next;
}
// Splice in a node containing the new element
pred.next = new Node(e, pred.next);
// Is there a new last element ?
if (pred.next.next == null)
last = pred.next;
}
/**
The toString method computes the string
representation of the list.
@return The string form of the list.
*/
public String toString()
{
StringBuilder strBuilder = new StringBuilder();
// Use p to walk down the linked list
Node p = first;
while (p != null)
{
strBuilder.append(p.value + \"\ \");
p = p.next;
}
return strBuilder.toString();
}
/**
The remove method removes the element at an index.
@param index The index of the element to remove.
@return The element removed.
@exception IndexOutOfBoundsException When index is
out of bounds.
*/
public String remove(int index)
{
if (index < 0 || index >= size())
{
String message = String.valueOf(index);
throw new IndexOutOfBoundsException(message);
}
String element; // The element to return
if (index == 0)
{
// Removal of first item in the list
element.
10-linked-list.ppt
This document summarizes the basics of linked lists, including how references to objects work in linked lists, traversing a linked list to access all nodes, and implementing common linked list operations like add, get, and add at a specific index. Key points covered include using a "current" reference to traverse the list without modifying the original reference, how to add nodes to an empty list versus a non-empty list, and how to insert nodes by modifying the next reference of the preceding node. Pseudocode is provided for linked list traversal and implementing the main linked list methods on a LinkedIntList class.
--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
This document contains code and comments for implementing a linked list data structure in C++. It includes function definitions and comments for common linked list operations like adding/removing nodes from the front/rear, checking if empty, getting length, converting to a string, and displaying all nodes. There are also TODO notes to fill in code for more advanced operations like deleting/inserting at a specified position, a copy constructor, and an assignment operator.
Linked list and its operations - Traversal
The document discusses linked lists and operations performed on singly linked lists. It defines a linked list as a data structure containing nodes that point to the next node in the list. Singly linked lists contain nodes with a data field and pointer to the next node. Common operations on singly linked lists include traversing the list, inserting and deleting nodes from different positions, searching for a node, sorting list elements, and merging two linked lists.
Linked list
Linked lists are linear data structures where each node contains a data field and a pointer to the next node. There are two types: singly linked lists where each node has a single next pointer, and doubly linked lists where each node has next and previous pointers. Common operations on linked lists include insertion and deletion which have O(1) time complexity for singly linked lists but require changing multiple pointers for doubly linked lists. Linked lists are useful when the number of elements is dynamic as they allow efficient insertions and deletions without shifting elements unlike arrays.
Recommended
This assignment and the next (#5) involve design and development of a.pdf
This assignment and the next (#5) involve design and development of a sequential
non contiguous and dynamic datastructure called LinkedList. A linked list object is
a container consisting of connected ListNode objects. As before, we are not going
to use pre-fabricated classes from the c++ library, but construct the LinkedList
ADT from scratch.
The first step is construction and testing of the ListNode class. A ListNode object
contains a data field and a pointer to the next ListNode object (note the recursive
definition).
#This assignment requires you to
1. Read the Assignment 4 Notes
2. Watch the Assignment 4 Support video
3. Implement the following methods of the ListNode class
-custom constructor
-setters for next pointer and data
4. Implement the insert and remove method in the main program
5. Scan the given template to find the above //TODO and implement the code
needed
//TODO in ListNodecpp.h file
public: ListNode(T idata, ListNode<T> * newNext);
public: void setNext(ListNode<T> * newNext);
public: void setData(T newData);
// TODO in main program
void remove(ListNode<int> * &front,int value)
void insert(ListNode<int> * &front,int value)
# The driver is given ListNodeMain.cpp is given to you that does the following
tests
1. Declares a pointer called front to point to a ListNode of datatype integer
2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked
list.
3. Inserts ListNode with data 3 to the list
4. Removes node 1 and adds it back to test removing and adding the first
element
5. Removes node 3 to test removing a middle node
6. Removes node 4 to test removing the last node
7. Attempt to remove a non existent node
8. Remove all existing nodes to empty the list
9. Insert node 4 and then node 1 to test if insertions preserve order
10.Print the list
Main.cpp
#include <iostream>
#include "ListNodecpp.h"
// REMEMBER each ListNode has two parts : a data field
// and an address field. The address is either null or points to the next node
//in the chain
//Requires: integer value for searching, address of front
//Effects: traverses the list node chain starting from front until the end comparing search value
with listnode getData. Returns the original search value if found, if not adds +1 to indicate not
found
//Modifies: Nothing
int search(ListNode<int> * front, int value);
//Requires: integer value for inserting, address of front
//Effects: creates a new ListNode with value and inserts in proper position (increasing order)in
the chain. If chain is empty, adds to the beginning
//Modifies: front, if node is added at the beginning.
//Also changes the next pointer of the previous node to point to the
//newly inserted list node. the next pointer of the newly inserted pointer
//points to what was the next of the previous node.
//This way both previous and current links are adjusted
//******** NOTE the use of & in passing pointer to front as parameter -
// Why do you think this is needed ?**********
void insert(ListNode<int> * &fr.
please i need help Im writing a program to test the merge sort alg.pdf
please i need help I\'m writing a program to test the merge sort algorithm for linked lists.
I need to Print the list before being sorted and the list after sorting the elements
here is my code only help in test program the four classes are ok except the test program.
TEST PROGRAM
import java.io.*;
import java.util.*;
public class Ch9_ProgExercise7 {
Scanner input = new Scanner(System.in);
public static void main(String[] args) throws IOException {
OrderedLinkedList list
= new OrderedLinkedList(); //Line 1
IntElement num = new IntElement(); //Line 2
// Prompt the user to enter some integers
System.out.println(\"Enter integers numbers \");
String temp = input.hasnextLine();
String[] split = temp.split(\" \");
for (int i = 0; i < split.length; i++) {
Integer a = Integer.parseInt(split[i]);
// insert the intgers into the list you created above
num.setNum(a);
list.insertNode(num);
}
System.out.println();
// Print the list before being sorted
// call the mergesort method
System.out.println();
// print the list after sorting the elements
}
}
ORDEREDLICKEDLIST CLASS
public class OrderedLinkedList extends LinkedListClass
{
//default constructor
public OrderedLinkedList()
{
super();
}
//copy constructor
public OrderedLinkedList(OrderedLinkedList otherList)
{
super(otherList);
}
public void linkedInsertionSort()
{
LinkedListNode lastInOrder;
LinkedListNode firstOutOfOrder;
LinkedListNode current;
LinkedListNode trailCurrent;
lastInOrder = first;
if(first == null)
System.out.println(\"Cannot sort an empty list\");
else
if(first.link == null)
System.out.println(\"The list is of length 1. \"
+ \"Already in order\");
else
while(lastInOrder.link != null)
{
firstOutOfOrder = lastInOrder.link;
if(firstOutOfOrder.info.compareTo(first.info) < 0)
{
lastInOrder.link = firstOutOfOrder.link;
firstOutOfOrder.link = first;
first = firstOutOfOrder;
}
else
{
trailCurrent = first;
current = first.link;
while(current.info.compareTo(firstOutOfOrder.info) < 0)
{
trailCurrent = current;
current = current.link;
}
if(current != firstOutOfOrder)
{
lastInOrder.link = firstOutOfOrder.link;
firstOutOfOrder.link = current;
trailCurrent.link = firstOutOfOrder;
}
else
lastInOrder = lastInOrder.link;
}
}
}//end linkedInsertionSort
//Method to determine whether searchItem is in
//the list.
//Postcondition: Returns true if searchItem is found
// in the list; otherwise, it returns
// false.
public boolean search(DataElement searchItem)
{
LinkedListNode current; //variable to traverse the list
boolean found;
current = first; //set current to point to the first
//node in the list
found = false; //set found to false
while(current != null && !found ) //search the list
if(current.info.compareTo(searchItem) >= 0)
found = true;
else
current = current.link; //make current point to
//the next node
if(found)
found = current.info.equals(searchItem);
return found;
}
//Method to insert insertItem in the list
//Postcondition: first points to the new list,
// newItem is inserted at the proper place
//.
The LinkedList1 class implements a Linked list. class.pdf
/**
The LinkedList1 class implements a Linked list.
*/
class LinkedList1
{
/**
The Node class stores a list element
and a reference to the next node.
*/
private class Node
{
String value;
Node next;
/**
Constructor.
@param val The element to store in the node.
@param n The reference to the successor node.
*/
Node(String val, Node n)
{
value = val;
next = n;
}
/**
Constructor.
@param val The element to store in the node.
*/
Node(String val)
{
// Call the other (sister) constructor.
this(val, null);
}
}
private Node first; // list head
private Node last; // last element in list
/**
Constructor.
*/
public LinkedList1()
{
first = null;
last = null;
}
/**
The isEmpty method checks to see
if the list is empty.
@return true if list is empty,
false otherwise.
*/
public boolean isEmpty()
{
return first == null;
}
/**
The size method returns the length of the list.
@return The number of elements in the list.
*/
public int size()
{
int count = 0;
Node p = first;
while (p != null)
{
// There is an element at p
count ++;
p = p.next;
}
return count;
}
/**
The add method adds an element to
the end of the list.
@param e The value to add to the
end of the list.
*/
public void add(String e)
{
if (isEmpty())
{
first = new Node(e);
last = first;
}
else
{
// Add to end of existing list
last.next = new Node(e);
last = last.next;
}
}
/**
The add method adds an element at a position.
@param e The element to add to the list.
@param index The position at which to add
the element.
@exception IndexOutOfBoundsException When
index is out of bounds.
*/
public void add(int index, String e)
{
if (index < 0 || index > size())
{
String message = String.valueOf(index);
throw new IndexOutOfBoundsException(message);
}
// Index is at least 0
if (index == 0)
{
// New element goes at beginning
first = new Node(e, first);
if (last == null)
last = first;
return;
}
// Set a reference pred to point to the node that
// will be the predecessor of the new node
Node pred = first;
for (int k = 1; k <= index - 1; k++)
{
pred = pred.next;
}
// Splice in a node containing the new element
pred.next = new Node(e, pred.next);
// Is there a new last element ?
if (pred.next.next == null)
last = pred.next;
}
/**
The toString method computes the string
representation of the list.
@return The string form of the list.
*/
public String toString()
{
StringBuilder strBuilder = new StringBuilder();
// Use p to walk down the linked list
Node p = first;
while (p != null)
{
strBuilder.append(p.value + \"\ \");
p = p.next;
}
return strBuilder.toString();
}
/**
The remove method removes the element at an index.
@param index The index of the element to remove.
@return The element removed.
@exception IndexOutOfBoundsException When index is
out of bounds.
*/
public String remove(int index)
{
if (index < 0 || index >= size())
{
String message = String.valueOf(index);
throw new IndexOutOfBoundsException(message);
}
String element; // The element to return
if (index == 0)
{
// Removal of first item in the list
element.
10-linked-list.ppt
This document summarizes the basics of linked lists, including how references to objects work in linked lists, traversing a linked list to access all nodes, and implementing common linked list operations like add, get, and add at a specific index. Key points covered include using a "current" reference to traverse the list without modifying the original reference, how to add nodes to an empty list versus a non-empty list, and how to insert nodes by modifying the next reference of the preceding node. Pseudocode is provided for linked list traversal and implementing the main linked list methods on a LinkedIntList class.
--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
This document contains code and comments for implementing a linked list data structure in C++. It includes function definitions and comments for common linked list operations like adding/removing nodes from the front/rear, checking if empty, getting length, converting to a string, and displaying all nodes. There are also TODO notes to fill in code for more advanced operations like deleting/inserting at a specified position, a copy constructor, and an assignment operator.
Linked list and its operations - Traversal
The document discusses linked lists and operations performed on singly linked lists. It defines a linked list as a data structure containing nodes that point to the next node in the list. Singly linked lists contain nodes with a data field and pointer to the next node. Common operations on singly linked lists include traversing the list, inserting and deleting nodes from different positions, searching for a node, sorting list elements, and merging two linked lists.
Linked list
Linked lists are linear data structures where each node contains a data field and a pointer to the next node. There are two types: singly linked lists where each node has a single next pointer, and doubly linked lists where each node has next and previous pointers. Common operations on linked lists include insertion and deletion which have O(1) time complexity for singly linked lists but require changing multiple pointers for doubly linked lists. Linked lists are useful when the number of elements is dynamic as they allow efficient insertions and deletions without shifting elements unlike arrays.
1#include stdio.h#include stdlib.h#include assert.h .pdf
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
2
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
Solution
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node*.
Linked List
Singly linked lists, doubly linked lists, and circular linked lists are types of linked data structures. Singly linked lists contain nodes with a data field and a next pointer. Doubly linked lists contain additional previous pointers. Circular linked lists connect the last node to the first node, making the list cyclic. Pseudocode is provided for common linked list operations like insertion, deletion, searching, on different positions in the lists.
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Mi 103 linked list
A linked list is a data structure where each node contains a data field and a pointer to the next node. Nodes can be dynamically added or removed, allowing the list to grow and shrink. The key operations on a linked list are appending a node to the end, inserting a node in order, and deleting a node. Each node is allocated dynamically and linked together via next pointers, allowing flexible and efficient insertion and removal of nodes anywhere in the list.
Template LinkedList;I am using templates to make some linkedLists.pdf
Sargent Corporation bought equipment on January 1, 2013. The equipment cost $180,000 and
had an expected salvage value of $30,000. The life of the equipment was estimated to be 6 years.
The book value of the equipment at the beginning of the third year using straight-line
depreciation would be . C $180,000. $150,000. $130,000. $50,000.
Solution
As per the information provided, Sargent Corporation bought an equipment on Jan 1, 2013 for
$180000 with 6 years life and salvage value of $30000.
Therefore, depreciation expense every year as per Straight line method would be:-
Equipment cost = $180000
LESS:- Salvage value = $30000
Amount on which depriciation is to be charged =$150000
Now, depreciation as per straight line method = $150000 / 6 years
= $25000 every year
Now, book value of Equipment at the beginning of third year is:-
Equipment cost = $180000
LESS :- Depreciation for first and second year = $25000 x 2 years = $50000
Therefore, book value at begginning of third year =$130000
Hence, as per the problem \"Option C\" is correct with book value of $130000..
I have been tasked to write a code for a Singly Linked list that inc.pdf
I have been tasked to write a code for a Singly Linked list that includes the following: Implement
a singly linked list class with the following methods: - `void insertAtBeginning(int value)`:
Inserts a new node with the given value at the beginning of the list. - `void insertAtEnd(int
value)`: Inserts a new node with the given value at the end of the list. - `void deleteNode(int
value)`: Deletes the first occurrence of a node with the given value. - `void display()`: Displays
the elements of the linked list. I have everything EXCEPT for the deleteNode. I have it included,
but it is throwing an error....I just need a little extra help to fix that part. My Code: #
#include
using namespace std;
class Node { //user defined type.
public: // public class modifier to access the class node.
int value;
Node* next;
};
void display(Node *head) {
if(head == NULL){
cout << "NULL" << endl;
}
else {
cout << head->value << endl;
display(head->next);
}
}
void insertAtBeginning(Node**head, int newValue){
// Steps to add a Node to the beginning of the linked list
//1. Prepare a newNode
Node * newNode = new Node();
newNode ->value = newValue;
//2. Put it in front of the current head
newNode->next = *head;
//3. Move head of the list to point to the newNode
*head= newNode;
}
void insertAtEnd(Node**head, int newValue) {
//1.Prepare a newNode
Node* newNode= new Node();
newNode ->value = newValue;
newNode->next = NULL;
//2. If Linked list is empty, newNode will be a head node
if(*head ==NULL){
*head = newNode;
return;
}
//3. Find last node
Node* last= *head;
while(last->next!=NULL){
last=last->next;
}
//4. Insert newNode after last node (at the end of List)
last->next =newNode;
}
void deleteNode(Node**before_del){
Node* temp;
temp = before_del->next;
before_del->next = temp->next;
delete temp;
}
int main()
{
Node* head=new Node();
Node* second=new Node();
Node* third=new Node();
Node* fourth=new Node();
head->value = 7;
head->next = second;
second->value = 12;
second->next = third;
third->value = 15;
third->next = fourth;
fourth->value = 20;
fourth->next = NULL;
insertAtBeginning(&head, 4);
insertAtBeginning(&head, 6);
insertAtEnd(&head, 23);
del(&head(15)->next);
display(head);
}.
C Exam Help
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please help me in C++Objective Create a singly linked list of num.pdf
please help me in C++
Objective: Create a singly linked list of numbers based upon user input.
Program logic: Ask for a number, add that number to the front of the list, print the list. Repeat
until they enter -1 for the number. .
Sample Input: 10, 15, 5, 2, 4, -1 Output: 4, 2, 5, 15, 10
Solution
#include
#include
using namespace std;
// It contains the value and the pointer to the next element in the list.
struct element {
int value;
struct element *next;
};
struct element *head = NULL;
// The list is stored in the object of this class.
class linkedlist {
public:
// To display list, we first see if head is NULL,
// If head is NULL it means the list is empty.
// Then we traverse along the list and display each item, until the next pointer is NULL.
void display() {
if(head == NULL){
cout << \"The list is empty\ \";
} else {
struct element *temp;
temp = head;
while(temp != NULL){
cout << temp->value << \" \" ;
temp = temp->next;
}
cout << \"\ \";
}
}
// To add an element to the list,create a newElement and store the value.
// If head is NULL, add the newElement to the head.
// To add an element at the starting of the list.
// The next pointer of the newElement must point to the old head.
// The new head must be pointed to the newElement.
void add(int a){
struct element *newElement;
newElement = (struct element *)malloc(sizeof(struct element));
newElement->value = a;
newElement->next = NULL;
if(head == NULL) {
head = newElement;
} else {
newElement->next = head;
head = newElement;
}
}
};
int main()
{
// Create an object of the class linkedlist.
linkedlist l = linkedlist();
int a;
do {
cout << \"Enter a number : \";
cin >> a;
if(a!=-1) { // If the number is -1 do not add it to the list.
l.add(a); // Add the number to the list.
}
l.display(); // Display the list.
} while(a!=-1); // Keep asking for a number until -1 is pressed.
return 0;
}.
Linked lists a
In computer science, a linked list is a linear collection of data elements, in which linear order is not given by their physical placement in memory. Instead, each element points to the next
linkedlist-130914084342-phpapp02.pptx
This document provides information about linked lists. It defines a linked list as a linear data structure where nodes are linked using pointers. Each node contains a data field and a pointer to the next node. It compares linked lists to arrays, noting that linked lists can dynamically allocate memory as needed, while arrays have fixed sizes. It describes the two main types of linked lists - singly linked and doubly linked. It provides examples of basic linked list operations like insertion, deletion and traversal for both singly and doubly linked lists. It also discusses some applications of linked lists like implementing stacks, queues and representing polynomials.
Data Structures in C++I am really new to C++, so links are really .pdf
Data Structures in C++
I am really new to C++, so links are really hard topic for me. It would be nice if you can provide
explanations of what doubly linked lists are and of some of you steps... Thank you
In this assignment, you will implement a doubly-linked list class, together with some list
operations. To make things easier, you’ll implement a list of int, rather than a template class.
Solution
A variable helps us to identify the data. For ex: int a = 5; Here 5 is identified through variable a.
Now, if we have collection of integers, we need some representation to identify them. We call it
array. For ex: int arr[5]
This array is nothing but a Data Structure.
So, a Data Structure is a way to group the data.
There are many Data Structures available like Arrays, Linked List, Doubly-Linked list, Stack,
Queue, etc.
Doubly-Linked list are the ones where you can traverse from the current node both in left and
right directions.
Why so many different types of Data Structures are required ?
Answer is very simple, grouping of data, storage of data and accessing the data is different.
For example, in case of Arrays we store all the data in contiguous locations.
What if we are not able to store the data in contiguous locations because we have huge data.
Answer is go for Linked List/Doubly-Linked list.
Here we can store the data anywhere and link the data through pointers.
I will try to provide comments for the code you have given. May be this can help you.
#pragma once
/*
dlist.h
Doubly-linked lists of ints
*/
#include
class dlist {
public:
dlist() { }
// Here we are creating a NODE, it has a integer value and two pointers.
// One pointer is to move to next node and other to go back to previous node.
struct node {
int value;
node* next;
node* prev;
};
// To return head pointer, i.e. start of the Doubly-Linked list.
node* head() const { return _head; }
// To return Tail pointer, i.e. end of the Doubly-Linked list.
node* tail() const { return _tail; }
// **** Implement ALL the following methods ****
// Returns the node at a particular index (0 is the head).
node* at(int index){
int cnt = 0;
struct node* tmp = head();
while(tmp!=NULL)
{
if (cnt+1 == index)
return tmp;
tmp = tmp->next;
}
}
// Insert a new value, after an existing one
void insert(node *previous, int value){
// check if the given previous is NULL
if (previous == NULL)
{
printf(\"the given previous node cannot be NULL\");
return;
}
// allocate new node
struct node* new_node =(struct node*) malloc(sizeof(struct node));
// put in the data
new_node->data = new_data;
// Make next of new node as next of previous
new_node->next = previous->next;
// Make the next of previous as new_node
previous->next = new_node;
// Make previous as previous of new_node
new_node->prev = previous;
// Change previous of new_node\'s next node
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
// Delete the given node
void del(node* which){
struct node* head_ref = head();
/* base case */
if(*head_ref == NUL.
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Let n = size of the hash table hashTable = array of size n, w.pdf
//Let n = size of the hash table
// hashTable = array of size n, which represents the hash table
// hashIndex = represents the index of array
// If the list is EMPTY, we add the element into the list
/*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such
a way to get sorted.*/
struct node
{
int key;
T value;
struct node *next;
};
struct node *hashTable[n] = {NULL};
struct node * createNode(int key, char *name, int age)
{
struct node *newnode;
newnode = (struct node *)malloc(sizeof(struct node));
newnode->key = key;
newnode->value = value;
newnode->next = NULL;
return newnode;
}
add(int key, T value)
{
int hashIndex = key % n;
struct node *newNode = creteNode(key,value);
/*Adding new node to the list with No element */
if (hashTable[hashIndex] == NULL)
{
hashTable[hashIndex] = newnode;
return;
}
/*Adding new node to the sorted list */
struct node *temp1;
struct node *temp2 = hashTable[hashIndex];
temp1 = temp2;
while(temp2 -> key < newNode -> key)
{
temp1 = temp2;
temp2 = temp2 ->next;
}
newnode->next = temp2;
temp1->next = newNode;
return;
}
Solution
//Let n = size of the hash table
// hashTable = array of size n, which represents the hash table
// hashIndex = represents the index of array
// If the list is EMPTY, we add the element into the list
/*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such
a way to get sorted.*/
struct node
{
int key;
T value;
struct node *next;
};
struct node *hashTable[n] = {NULL};
struct node * createNode(int key, char *name, int age)
{
struct node *newnode;
newnode = (struct node *)malloc(sizeof(struct node));
newnode->key = key;
newnode->value = value;
newnode->next = NULL;
return newnode;
}
add(int key, T value)
{
int hashIndex = key % n;
struct node *newNode = creteNode(key,value);
/*Adding new node to the list with No element */
if (hashTable[hashIndex] == NULL)
{
hashTable[hashIndex] = newnode;
return;
}
/*Adding new node to the sorted list */
struct node *temp1;
struct node *temp2 = hashTable[hashIndex];
temp1 = temp2;
while(temp2 -> key < newNode -> key)
{
temp1 = temp2;
temp2 = temp2 ->next;
}
newnode->next = temp2;
temp1->next = newNode;
return;
}.
double Skiplist code below#include iostream#include vector.pdf
double Skiplist code below
#include
#include
#include
#include
#include
#include
#include "Node.h"
#ifndef SKIPLIST_H
#define SKIPLIST_H
using namespace std;
class SkipList{
public:
// Maximum number of levels allowed for this skip list
int MAXIMUM_ALLOWED_LEVELS;
// current maximum level amongst the inserted nodes
int level;
// the head node's next links are connected to the first node at every level
Node *head;
SkipList(int maxLevels){
MAXIMUM_ALLOWED_LEVELS = maxLevels;
// initially we have the bottom-most level only
level = 0;
// create the header node, value is irrelevant (as long as it doesn't match an inserted value -
NO REPEATS), number of next links is important (this node is connected to the first node at
every level)
head = new Node(INT_MIN, MAXIMUM_ALLOWED_LEVELS);
}
int RandomLevel(){
float probablity = (float)rand()/RAND_MAX; // flip a coin
int lvl = 0;
/*Olog(n) */while (probablity < 0.5 && lvl < MAXIMUM_ALLOWED_LEVELS){
lvl++; // landed heads so increase level by 1
probablity = (float)rand()/RAND_MAX; // flip a coin again
}
return lvl;
}
Node* CreateNode(int value, int level){
// create a new node with next links for every level that this node will be part of
// it will use these links to connect to the next node at each level
return new Node(value, level);
}
void InsertElement(int value)/*Olog(n) becuase all the functions in it are Olog(n) */{
/**
* THIS CODE HAS A BUG,
*
* ***********FIND THE BUG,
* *************EXPLAIN THE WHY THE BUG EXIST
* AND
* **********************SUGGEST A FIX *
*
*/
//w
Node *current = head; // start at head node at the toppest level
/*Olog(n)*/vector update(MAXIMUM_ALLOWED_LEVELS+1, nullptr); // this will hold
the nodes that need updating at every level after the insert takes place
/*BIG Olog(n)*/for (int i = level; i >= 0; i--){ //level is the current maximum level
// for level i, if value is to be inserted here then find out where (i.e. after which node)
/*O(1) constant */ while (current->next[i] != nullptr && current->next[i]->value <
value){ //the while loop is saying keep moving right until null or the current.value > value
current = current->next[i];
}
// found the node after which the value is to be placed at level i
update[i] = current;
}
//if
current = current->next[0]; // current moves bcoz we need to check if the value exists
if (current == nullptr || current->value != value){
/*Olog(n) */int ranLevel = RandomLevel(); // choose a random level upto where the new
node will be placed
if (ranLevel > level){
// if random level for current node is higher than the current maximum level for the
existing nodes then set updates for each level over current maximum to head (since there are no
nodes at these levels yet).
/*Olog(n)*/for (int i=level+1;i<=ranLevel;i++){
update[i] = head;
}
// also change the current maximum level for the existing nodes
level = ranLevel;
}
// create new node with next links for every level from ranLevel to zero
/*Olog(n) */Node* n = CreateNode(value, ranLevel);
// placi.
Data Structures - Lecture 7 [Linked List]
- A linked list is a data structure where each node contains a data field and a pointer to the next node.
- It allows dynamic size and efficient insertion/deletion compared to arrays.
- A doubly linked list adds a pointer to the previous node, allowing traversal in both directions.
- A circular linked list connects the last node back to the first node, making it a continuous loop.
- Variations require changes to the node structure and functions like append/delete to handle the added previous/next pointers.
Link list part 1
The document discusses linked lists and their implementation in C++. It covers key concepts like nodes, pointers, traversal, insertion, deletion and searching in a linked list. It provides pseudocode and C++ code for functions to insert and delete nodes from the head of the list, traverse the entire list, and search for a node containing a given data value.
Dividing a linked list into two sublists of almost equal sizesa. A.pdf
Dividing a linked list into two sublists of almost equal sizes
a. Add the operation divideMid to the class linkedListType as follows:
Consider the following statements:
unorderedLinkedList myList;
unorderedLinkedList subList;
Suppose myList points to the list with elements 34 65 27 89 12 (in this order). The statement:
myList.divideMid(subList); divides myList into two sublists: myList points to the list with the
elements 34 65 27,
and subList points to the sublist with the elements 89 12.
b. Write the definition of the function template to implement the operation divideMid. Also,
write a program to test your function. The header files linkedList.h and unorderedLinkedList.h
are supplied. Your test program should produce output similar to this:
Turn in:
linkedList.h with modifications
Your test program.
(C++)
Required code is below:
#ifndef H_UnorderedLinkedList
#define H_UnorderedLinkedList
#include \"linkedList.h\"
using namespace std;
template
class unorderedLinkedList: public linkedListType
{
public:
bool search(const Type& searchItem) const;
//Function to determine whether searchItem is in the list.
//Postcondition: Returns true if searchItem is in the
// list, otherwise the value false is
// returned.
void insertFirst(const Type& newItem);
//Function to insert newItem at the beginning of the list.
//Postcondition: first points to the new list, newItem is
// inserted at the beginning of the list,
// last points to the last node in the
// list, and count is incremented by 1.
void insertLast(const Type& newItem);
//Function to insert newItem at the end of the list.
//Postcondition: first points to the new list, newItem
// is inserted at the end of the list,
// last points to the last node in the
// list, and count is incremented by 1.
void deleteNode(const Type& deleteItem);
//Function to delete deleteItem from the list.
//Postcondition: If found, the node containing
// deleteItem is deleted from the list.
// first points to the first node, last
// points to the last node of the updated
// list, and count is decremented by 1.
};
template
bool unorderedLinkedList::
search(const Type& searchItem) const
{
nodeType *current; //pointer to traverse the list
bool found = false;
current = this->first; //set current to point to the first
//node in the list
while (current != NULL && !found) //search the list
if (current->info == searchItem) //searchItem is found
found = true;
else
current = current->link; //make current point to
//the next node
return found;
}//end search
template
void unorderedLinkedList::insertFirst(const Type& newItem)
{
nodeType *newNode; //pointer to create the new node
newNode = new nodeType; //create the new node
newNode->info = newItem; //store the new item in the node
newNode->link = this->first; //insert newNode before first
this->first = newNode; //make first point to the
//actual first node
this->count++; //increment count
if (this->last == NULL) //if the list was empty, newNode is also
//the last node in the list
this->l.
Lec-4_Linked-List (1).pdf
A linked list is a data structure made up of nodes that are connected to each other. Each node contains data and a link to the next node. Linked lists can grow and shrink dynamically as nodes are added or removed. There are several types of linked lists including single linked lists where navigation is forward only, doubly linked lists where navigation is bidirectional, and circular linked lists where the last node links back to the first node. Common operations on linked lists include appending nodes to add to the end, inserting nodes in a sorted manner, and deleting nodes by value. These operations involve allocating memory for new nodes, linking nodes together, and removing nodes by adjusting pointers.
using the single linked list code written in the class or in the lab,.pdf
using the single linked list code written in the class or in the lab, write and test the follow
function: Merge() function that takes two lists, each of which is sorted in increasing order, and
merges the two together into one list which is in increasing order.
Solution
//The main class
class MergeList {
Node head; // head of list
static Node a, b;
//Node class
static class Node {
int element;
Node next;
// Constructor to create a new node
Node(int d) {
element = d;
next = null;
}
}
void printlist(Node node) {
while (node != null) {
System.out.print(node.element + \" \");
node = node.next;
}
}
Node sortMerge(Node node1, Node node2) {
// Here we will check whether both the list are null or not.
if (node1 == null && node2 == null) {
return null;
}
// sortResultant node
Node sortRes = null;
// Now we will compare and merge the elements of both the list.
while (node1 != null && node2 != null) {
// We will compare current element of both lists
if (node1.element >= node2.element) {
Node temp = node1.next;
node1.next = sortRes;
sortRes = node1;
node1 = temp;
} else {
Node temp = node2.next;
node2.next = sortRes;
sortRes = node2;
node2 = temp;
}
}
// If second list reached end, but first list has
// nodes. Add remaining nodes of first list at the
// front of sortResult list
while (node1 != null) {
Node temp = node1.next;
node1.next = sortRes;
sortRes = node1;
node1 = temp;
}
// If first list reached end, but second list has
// node. Add remaining nodes of first list at the
// front of sortResult list
while (node2 != null) {
Node temp = node2.next;
node2.next = sortRes;
sortRes = node2;
node2 = temp;
}
return sortRes;
}
public static void main(String[] args) {
MergeList list = new MergeList();
Node sortResult = null;
//Create two linked list
list.a = new Node(5);
list.a.next = new Node(10);
list.a.next.next = new Node(15);
list.b = new Node(2);
list.b.next = new Node(3);
list.b.next.next = new Node(20);
System.out.println(\"List a before merge :\");
list.printlist(a);
System.out.println(\"\");
System.out.println(\"List b before merge :\");
list.printlist(b);
// Cal the function to merge the list in increasing order.
sortResult = list.sortedmerge(a, b);
System.out.println(\"\");
System.out.println(\"Merged linked list : \");
list.printlist(sortResult);
}
}.
Homework 05 - Linked Lists (C++)(1) Implement the concepts of a un.pdf
Homework 05 - Linked Lists (C++)
(1) Implement the concepts of a union, and difference as defined noted in Chapter 1 questions 6,
and 8 using a linked list. You should not use arrays, dynamic arrays or the STL vector or STL
list classes.
You will be linking a series of nodes together to implement a single linked list class. Define a
Node as a struct that holds a character (like they did in the book) and a pointer to another Node:
struct Node {
char data;
Node* next;
};
Then define a LinkedList class which has (at a minimum) these member functions:
LinkedList();
~LinkedList();
bool insertAtFront();
bool insertBeforePosition(int index);
bool insertAtBack();
bool deleteAtFront();
bool deleteBeforePosition(int index);
bool deleteAtBack();
Implement each of these functions to provide the correct functionality. These functions return
true if successful in inserting or deleting, otherwise they return false (indicating the operation
was not successful).
Finally, create an overloaded + operator to handle the Union of two linked lists and an
overloaded - operator to handle the Difference of two linked lists.
Because we are dealing with pointers you should have both a LinkedList Constructor and
Destructor. Remember that you do not directly call a Constructor or Destructor Function. The
Destructor is automatically called when the variable loses scope or the program ends.
Remember, that we are dealing with not just one dynamically allocated Node (with the new
operator), but many, so you will have to start at the head of the list and go until the Node points
to nullptr. Then keep deleting the previous Node pointer until there are no Nodes left to delete.
(2) To verify your set class, write a main function that takes two lines of characters from the
input file input.txt and store each line of characters into two separate linked lists. Then using
these two lists, perform the Union and set Difference of the characters from the file, and print the
results to the console.
(3) Please also complete an asymptotic (Big O) analysis of your insertAtFront() and
insertAtBack() member functions. Place this in a file called analysis.txt.
Solution
Here is the complete code for your question. Please do rate the aswer if you are happy with
program. Please read comments in the program. Also pay special attention to comments in + and
- operators. You WILL NEED to comment out if condition in + if you want all elements in both
lists irrespective of if they duplicates or no. For now implemented to have true behaviour of no
duplicates in union.
check the functions by creating some lists and playing around different fucntions like its done in
main and commented out.
As far as complexity:
for insertAtFront() its O(1) since it does not depend on the size of the list. It just replaces the start
of the list. Hence it is O(1) and independent of the size of list.
for insertAtBack(), its O(n) since it depends on the number of elements in the list. It should
traverse till end of list and .
4r = 2 L2r = L2thats none of the above, since you didnt wrote t.pdf
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2
Solution
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2.
41.Ans. Ans,Sarcoplasmic reticulumSarcoplasmic reticulum is a type.pdf
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
volume is cytosol.
the other options are wrong. membrane bound organelles have outer layer plasma membrane, so
the endoplasmic reticulum not far exceeds the area of plasma membrane. not only the nucleus
surrounded by double layer, other organells for example Mitochondria also surrounded by
double layers. and not only the nucleus, mitochondria also have the DNA. and chloroplast also
have the DNA.
43.Ans. True
once loaded vesicles are carry the cargo to different location with in the body, and reaches the
site of secretion. once reached some of the vesicles containing material for constitutive -
Produced in constant amount, releases and immediately fused with Plasma membrane. where as
vesicles called secretory vesicles- these are the vesicles in which molecules destined for
secretion are stored, these vesicles are in regulated pathway means once reached to the site of
secretion these vesicles wait at he membrane until it receives a signal to secrete, once cell
receives the signal these are going secrete the content and material fuse with the plasma
membrane
44.Ans.False
Nuclear localization signal needed to the peptide until it crosses the membrane
45Ans.True
Chaperones are needed for the import of the proteins. these are prevent the premature folding of
the proteins. while import into the mitochondria, the Chaperones present in the inner surface of
the membrane of mitochondria able to bind the protruding ends of polypeptide and prevents them
to diffuse back into the cytosol through the pore and carry them further and into the matrix. while
moving the peptides further into the matrix at every stage the se peptides are bounded with the
Chaperones which prevent backward movement of the peptides. this kind of diffusion called as
biased diffusion and Chaperones acted as Brownian ratchet.
46.Ans.E) All of These happen.
the above all happens when haploid cell mates
Solution
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
vol.
More Related Content
Similar to code ----- public void add(int key, T value) { Link previou.pdf
1#include stdio.h#include stdlib.h#include assert.h .pdf
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
2
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
Solution
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node*.
Linked List
Singly linked lists, doubly linked lists, and circular linked lists are types of linked data structures. Singly linked lists contain nodes with a data field and a next pointer. Doubly linked lists contain additional previous pointers. Circular linked lists connect the last node to the first node, making the list cyclic. Pseudocode is provided for common linked list operations like insertion, deletion, searching, on different positions in the lists.
C Homework Help
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You can also call on +1 678 648 4277 for any assistance with C Homework.
Mi 103 linked list
A linked list is a data structure where each node contains a data field and a pointer to the next node. Nodes can be dynamically added or removed, allowing the list to grow and shrink. The key operations on a linked list are appending a node to the end, inserting a node in order, and deleting a node. Each node is allocated dynamically and linked together via next pointers, allowing flexible and efficient insertion and removal of nodes anywhere in the list.
Template LinkedList;I am using templates to make some linkedLists.pdf
Sargent Corporation bought equipment on January 1, 2013. The equipment cost $180,000 and
had an expected salvage value of $30,000. The life of the equipment was estimated to be 6 years.
The book value of the equipment at the beginning of the third year using straight-line
depreciation would be . C $180,000. $150,000. $130,000. $50,000.
Solution
As per the information provided, Sargent Corporation bought an equipment on Jan 1, 2013 for
$180000 with 6 years life and salvage value of $30000.
Therefore, depreciation expense every year as per Straight line method would be:-
Equipment cost = $180000
LESS:- Salvage value = $30000
Amount on which depriciation is to be charged =$150000
Now, depreciation as per straight line method = $150000 / 6 years
= $25000 every year
Now, book value of Equipment at the beginning of third year is:-
Equipment cost = $180000
LESS :- Depreciation for first and second year = $25000 x 2 years = $50000
Therefore, book value at begginning of third year =$130000
Hence, as per the problem \"Option C\" is correct with book value of $130000..
I have been tasked to write a code for a Singly Linked list that inc.pdf
I have been tasked to write a code for a Singly Linked list that includes the following: Implement
a singly linked list class with the following methods: - `void insertAtBeginning(int value)`:
Inserts a new node with the given value at the beginning of the list. - `void insertAtEnd(int
value)`: Inserts a new node with the given value at the end of the list. - `void deleteNode(int
value)`: Deletes the first occurrence of a node with the given value. - `void display()`: Displays
the elements of the linked list. I have everything EXCEPT for the deleteNode. I have it included,
but it is throwing an error....I just need a little extra help to fix that part. My Code: #
#include
using namespace std;
class Node { //user defined type.
public: // public class modifier to access the class node.
int value;
Node* next;
};
void display(Node *head) {
if(head == NULL){
cout << "NULL" << endl;
}
else {
cout << head->value << endl;
display(head->next);
}
}
void insertAtBeginning(Node**head, int newValue){
// Steps to add a Node to the beginning of the linked list
//1. Prepare a newNode
Node * newNode = new Node();
newNode ->value = newValue;
//2. Put it in front of the current head
newNode->next = *head;
//3. Move head of the list to point to the newNode
*head= newNode;
}
void insertAtEnd(Node**head, int newValue) {
//1.Prepare a newNode
Node* newNode= new Node();
newNode ->value = newValue;
newNode->next = NULL;
//2. If Linked list is empty, newNode will be a head node
if(*head ==NULL){
*head = newNode;
return;
}
//3. Find last node
Node* last= *head;
while(last->next!=NULL){
last=last->next;
}
//4. Insert newNode after last node (at the end of List)
last->next =newNode;
}
void deleteNode(Node**before_del){
Node* temp;
temp = before_del->next;
before_del->next = temp->next;
delete temp;
}
int main()
{
Node* head=new Node();
Node* second=new Node();
Node* third=new Node();
Node* fourth=new Node();
head->value = 7;
head->next = second;
second->value = 12;
second->next = third;
third->value = 15;
third->next = fourth;
fourth->value = 20;
fourth->next = NULL;
insertAtBeginning(&head, 4);
insertAtBeginning(&head, 6);
insertAtEnd(&head, 23);
del(&head(15)->next);
display(head);
}.
C Exam Help
I am Gabriel C. I am a C Exam Expert at programmingexamhelp.com. I hold a PhD. in Business analyst of Information Technology, Montreal College of Information Technology, Canada. I have been helping students with their exams for the past 8 years. You can hire me to take your exam in C.
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please help me in C++Objective Create a singly linked list of num.pdf
please help me in C++
Objective: Create a singly linked list of numbers based upon user input.
Program logic: Ask for a number, add that number to the front of the list, print the list. Repeat
until they enter -1 for the number. .
Sample Input: 10, 15, 5, 2, 4, -1 Output: 4, 2, 5, 15, 10
Solution
#include
#include
using namespace std;
// It contains the value and the pointer to the next element in the list.
struct element {
int value;
struct element *next;
};
struct element *head = NULL;
// The list is stored in the object of this class.
class linkedlist {
public:
// To display list, we first see if head is NULL,
// If head is NULL it means the list is empty.
// Then we traverse along the list and display each item, until the next pointer is NULL.
void display() {
if(head == NULL){
cout << \"The list is empty\ \";
} else {
struct element *temp;
temp = head;
while(temp != NULL){
cout << temp->value << \" \" ;
temp = temp->next;
}
cout << \"\ \";
}
}
// To add an element to the list,create a newElement and store the value.
// If head is NULL, add the newElement to the head.
// To add an element at the starting of the list.
// The next pointer of the newElement must point to the old head.
// The new head must be pointed to the newElement.
void add(int a){
struct element *newElement;
newElement = (struct element *)malloc(sizeof(struct element));
newElement->value = a;
newElement->next = NULL;
if(head == NULL) {
head = newElement;
} else {
newElement->next = head;
head = newElement;
}
}
};
int main()
{
// Create an object of the class linkedlist.
linkedlist l = linkedlist();
int a;
do {
cout << \"Enter a number : \";
cin >> a;
if(a!=-1) { // If the number is -1 do not add it to the list.
l.add(a); // Add the number to the list.
}
l.display(); // Display the list.
} while(a!=-1); // Keep asking for a number until -1 is pressed.
return 0;
}.
Linked lists a
In computer science, a linked list is a linear collection of data elements, in which linear order is not given by their physical placement in memory. Instead, each element points to the next
linkedlist-130914084342-phpapp02.pptx
This document provides information about linked lists. It defines a linked list as a linear data structure where nodes are linked using pointers. Each node contains a data field and a pointer to the next node. It compares linked lists to arrays, noting that linked lists can dynamically allocate memory as needed, while arrays have fixed sizes. It describes the two main types of linked lists - singly linked and doubly linked. It provides examples of basic linked list operations like insertion, deletion and traversal for both singly and doubly linked lists. It also discusses some applications of linked lists like implementing stacks, queues and representing polynomials.
Data Structures in C++I am really new to C++, so links are really .pdf
Data Structures in C++
I am really new to C++, so links are really hard topic for me. It would be nice if you can provide
explanations of what doubly linked lists are and of some of you steps... Thank you
In this assignment, you will implement a doubly-linked list class, together with some list
operations. To make things easier, you’ll implement a list of int, rather than a template class.
Solution
A variable helps us to identify the data. For ex: int a = 5; Here 5 is identified through variable a.
Now, if we have collection of integers, we need some representation to identify them. We call it
array. For ex: int arr[5]
This array is nothing but a Data Structure.
So, a Data Structure is a way to group the data.
There are many Data Structures available like Arrays, Linked List, Doubly-Linked list, Stack,
Queue, etc.
Doubly-Linked list are the ones where you can traverse from the current node both in left and
right directions.
Why so many different types of Data Structures are required ?
Answer is very simple, grouping of data, storage of data and accessing the data is different.
For example, in case of Arrays we store all the data in contiguous locations.
What if we are not able to store the data in contiguous locations because we have huge data.
Answer is go for Linked List/Doubly-Linked list.
Here we can store the data anywhere and link the data through pointers.
I will try to provide comments for the code you have given. May be this can help you.
#pragma once
/*
dlist.h
Doubly-linked lists of ints
*/
#include
class dlist {
public:
dlist() { }
// Here we are creating a NODE, it has a integer value and two pointers.
// One pointer is to move to next node and other to go back to previous node.
struct node {
int value;
node* next;
node* prev;
};
// To return head pointer, i.e. start of the Doubly-Linked list.
node* head() const { return _head; }
// To return Tail pointer, i.e. end of the Doubly-Linked list.
node* tail() const { return _tail; }
// **** Implement ALL the following methods ****
// Returns the node at a particular index (0 is the head).
node* at(int index){
int cnt = 0;
struct node* tmp = head();
while(tmp!=NULL)
{
if (cnt+1 == index)
return tmp;
tmp = tmp->next;
}
}
// Insert a new value, after an existing one
void insert(node *previous, int value){
// check if the given previous is NULL
if (previous == NULL)
{
printf(\"the given previous node cannot be NULL\");
return;
}
// allocate new node
struct node* new_node =(struct node*) malloc(sizeof(struct node));
// put in the data
new_node->data = new_data;
// Make next of new node as next of previous
new_node->next = previous->next;
// Make the next of previous as new_node
previous->next = new_node;
// Make previous as previous of new_node
new_node->prev = previous;
// Change previous of new_node\'s next node
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
// Delete the given node
void del(node* which){
struct node* head_ref = head();
/* base case */
if(*head_ref == NUL.
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Let n = size of the hash table hashTable = array of size n, w.pdf
//Let n = size of the hash table
// hashTable = array of size n, which represents the hash table
// hashIndex = represents the index of array
// If the list is EMPTY, we add the element into the list
/*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such
a way to get sorted.*/
struct node
{
int key;
T value;
struct node *next;
};
struct node *hashTable[n] = {NULL};
struct node * createNode(int key, char *name, int age)
{
struct node *newnode;
newnode = (struct node *)malloc(sizeof(struct node));
newnode->key = key;
newnode->value = value;
newnode->next = NULL;
return newnode;
}
add(int key, T value)
{
int hashIndex = key % n;
struct node *newNode = creteNode(key,value);
/*Adding new node to the list with No element */
if (hashTable[hashIndex] == NULL)
{
hashTable[hashIndex] = newnode;
return;
}
/*Adding new node to the sorted list */
struct node *temp1;
struct node *temp2 = hashTable[hashIndex];
temp1 = temp2;
while(temp2 -> key < newNode -> key)
{
temp1 = temp2;
temp2 = temp2 ->next;
}
newnode->next = temp2;
temp1->next = newNode;
return;
}
Solution
//Let n = size of the hash table
// hashTable = array of size n, which represents the hash table
// hashIndex = represents the index of array
// If the list is EMPTY, we add the element into the list
/*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such
a way to get sorted.*/
struct node
{
int key;
T value;
struct node *next;
};
struct node *hashTable[n] = {NULL};
struct node * createNode(int key, char *name, int age)
{
struct node *newnode;
newnode = (struct node *)malloc(sizeof(struct node));
newnode->key = key;
newnode->value = value;
newnode->next = NULL;
return newnode;
}
add(int key, T value)
{
int hashIndex = key % n;
struct node *newNode = creteNode(key,value);
/*Adding new node to the list with No element */
if (hashTable[hashIndex] == NULL)
{
hashTable[hashIndex] = newnode;
return;
}
/*Adding new node to the sorted list */
struct node *temp1;
struct node *temp2 = hashTable[hashIndex];
temp1 = temp2;
while(temp2 -> key < newNode -> key)
{
temp1 = temp2;
temp2 = temp2 ->next;
}
newnode->next = temp2;
temp1->next = newNode;
return;
}.
double Skiplist code below#include iostream#include vector.pdf
double Skiplist code below
#include
#include
#include
#include
#include
#include
#include "Node.h"
#ifndef SKIPLIST_H
#define SKIPLIST_H
using namespace std;
class SkipList{
public:
// Maximum number of levels allowed for this skip list
int MAXIMUM_ALLOWED_LEVELS;
// current maximum level amongst the inserted nodes
int level;
// the head node's next links are connected to the first node at every level
Node *head;
SkipList(int maxLevels){
MAXIMUM_ALLOWED_LEVELS = maxLevels;
// initially we have the bottom-most level only
level = 0;
// create the header node, value is irrelevant (as long as it doesn't match an inserted value -
NO REPEATS), number of next links is important (this node is connected to the first node at
every level)
head = new Node(INT_MIN, MAXIMUM_ALLOWED_LEVELS);
}
int RandomLevel(){
float probablity = (float)rand()/RAND_MAX; // flip a coin
int lvl = 0;
/*Olog(n) */while (probablity < 0.5 && lvl < MAXIMUM_ALLOWED_LEVELS){
lvl++; // landed heads so increase level by 1
probablity = (float)rand()/RAND_MAX; // flip a coin again
}
return lvl;
}
Node* CreateNode(int value, int level){
// create a new node with next links for every level that this node will be part of
// it will use these links to connect to the next node at each level
return new Node(value, level);
}
void InsertElement(int value)/*Olog(n) becuase all the functions in it are Olog(n) */{
/**
* THIS CODE HAS A BUG,
*
* ***********FIND THE BUG,
* *************EXPLAIN THE WHY THE BUG EXIST
* AND
* **********************SUGGEST A FIX *
*
*/
//w
Node *current = head; // start at head node at the toppest level
/*Olog(n)*/vector update(MAXIMUM_ALLOWED_LEVELS+1, nullptr); // this will hold
the nodes that need updating at every level after the insert takes place
/*BIG Olog(n)*/for (int i = level; i >= 0; i--){ //level is the current maximum level
// for level i, if value is to be inserted here then find out where (i.e. after which node)
/*O(1) constant */ while (current->next[i] != nullptr && current->next[i]->value <
value){ //the while loop is saying keep moving right until null or the current.value > value
current = current->next[i];
}
// found the node after which the value is to be placed at level i
update[i] = current;
}
//if
current = current->next[0]; // current moves bcoz we need to check if the value exists
if (current == nullptr || current->value != value){
/*Olog(n) */int ranLevel = RandomLevel(); // choose a random level upto where the new
node will be placed
if (ranLevel > level){
// if random level for current node is higher than the current maximum level for the
existing nodes then set updates for each level over current maximum to head (since there are no
nodes at these levels yet).
/*Olog(n)*/for (int i=level+1;i<=ranLevel;i++){
update[i] = head;
}
// also change the current maximum level for the existing nodes
level = ranLevel;
}
// create new node with next links for every level from ranLevel to zero
/*Olog(n) */Node* n = CreateNode(value, ranLevel);
// placi.
Data Structures - Lecture 7 [Linked List]
- A linked list is a data structure where each node contains a data field and a pointer to the next node.
- It allows dynamic size and efficient insertion/deletion compared to arrays.
- A doubly linked list adds a pointer to the previous node, allowing traversal in both directions.
- A circular linked list connects the last node back to the first node, making it a continuous loop.
- Variations require changes to the node structure and functions like append/delete to handle the added previous/next pointers.
Link list part 1
The document discusses linked lists and their implementation in C++. It covers key concepts like nodes, pointers, traversal, insertion, deletion and searching in a linked list. It provides pseudocode and C++ code for functions to insert and delete nodes from the head of the list, traverse the entire list, and search for a node containing a given data value.
Dividing a linked list into two sublists of almost equal sizesa. A.pdf
Dividing a linked list into two sublists of almost equal sizes
a. Add the operation divideMid to the class linkedListType as follows:
Consider the following statements:
unorderedLinkedList myList;
unorderedLinkedList subList;
Suppose myList points to the list with elements 34 65 27 89 12 (in this order). The statement:
myList.divideMid(subList); divides myList into two sublists: myList points to the list with the
elements 34 65 27,
and subList points to the sublist with the elements 89 12.
b. Write the definition of the function template to implement the operation divideMid. Also,
write a program to test your function. The header files linkedList.h and unorderedLinkedList.h
are supplied. Your test program should produce output similar to this:
Turn in:
linkedList.h with modifications
Your test program.
(C++)
Required code is below:
#ifndef H_UnorderedLinkedList
#define H_UnorderedLinkedList
#include \"linkedList.h\"
using namespace std;
template
class unorderedLinkedList: public linkedListType
{
public:
bool search(const Type& searchItem) const;
//Function to determine whether searchItem is in the list.
//Postcondition: Returns true if searchItem is in the
// list, otherwise the value false is
// returned.
void insertFirst(const Type& newItem);
//Function to insert newItem at the beginning of the list.
//Postcondition: first points to the new list, newItem is
// inserted at the beginning of the list,
// last points to the last node in the
// list, and count is incremented by 1.
void insertLast(const Type& newItem);
//Function to insert newItem at the end of the list.
//Postcondition: first points to the new list, newItem
// is inserted at the end of the list,
// last points to the last node in the
// list, and count is incremented by 1.
void deleteNode(const Type& deleteItem);
//Function to delete deleteItem from the list.
//Postcondition: If found, the node containing
// deleteItem is deleted from the list.
// first points to the first node, last
// points to the last node of the updated
// list, and count is decremented by 1.
};
template
bool unorderedLinkedList::
search(const Type& searchItem) const
{
nodeType *current; //pointer to traverse the list
bool found = false;
current = this->first; //set current to point to the first
//node in the list
while (current != NULL && !found) //search the list
if (current->info == searchItem) //searchItem is found
found = true;
else
current = current->link; //make current point to
//the next node
return found;
}//end search
template
void unorderedLinkedList::insertFirst(const Type& newItem)
{
nodeType *newNode; //pointer to create the new node
newNode = new nodeType; //create the new node
newNode->info = newItem; //store the new item in the node
newNode->link = this->first; //insert newNode before first
this->first = newNode; //make first point to the
//actual first node
this->count++; //increment count
if (this->last == NULL) //if the list was empty, newNode is also
//the last node in the list
this->l.
Lec-4_Linked-List (1).pdf
A linked list is a data structure made up of nodes that are connected to each other. Each node contains data and a link to the next node. Linked lists can grow and shrink dynamically as nodes are added or removed. There are several types of linked lists including single linked lists where navigation is forward only, doubly linked lists where navigation is bidirectional, and circular linked lists where the last node links back to the first node. Common operations on linked lists include appending nodes to add to the end, inserting nodes in a sorted manner, and deleting nodes by value. These operations involve allocating memory for new nodes, linking nodes together, and removing nodes by adjusting pointers.
using the single linked list code written in the class or in the lab,.pdf
using the single linked list code written in the class or in the lab, write and test the follow
function: Merge() function that takes two lists, each of which is sorted in increasing order, and
merges the two together into one list which is in increasing order.
Solution
//The main class
class MergeList {
Node head; // head of list
static Node a, b;
//Node class
static class Node {
int element;
Node next;
// Constructor to create a new node
Node(int d) {
element = d;
next = null;
}
}
void printlist(Node node) {
while (node != null) {
System.out.print(node.element + \" \");
node = node.next;
}
}
Node sortMerge(Node node1, Node node2) {
// Here we will check whether both the list are null or not.
if (node1 == null && node2 == null) {
return null;
}
// sortResultant node
Node sortRes = null;
// Now we will compare and merge the elements of both the list.
while (node1 != null && node2 != null) {
// We will compare current element of both lists
if (node1.element >= node2.element) {
Node temp = node1.next;
node1.next = sortRes;
sortRes = node1;
node1 = temp;
} else {
Node temp = node2.next;
node2.next = sortRes;
sortRes = node2;
node2 = temp;
}
}
// If second list reached end, but first list has
// nodes. Add remaining nodes of first list at the
// front of sortResult list
while (node1 != null) {
Node temp = node1.next;
node1.next = sortRes;
sortRes = node1;
node1 = temp;
}
// If first list reached end, but second list has
// node. Add remaining nodes of first list at the
// front of sortResult list
while (node2 != null) {
Node temp = node2.next;
node2.next = sortRes;
sortRes = node2;
node2 = temp;
}
return sortRes;
}
public static void main(String[] args) {
MergeList list = new MergeList();
Node sortResult = null;
//Create two linked list
list.a = new Node(5);
list.a.next = new Node(10);
list.a.next.next = new Node(15);
list.b = new Node(2);
list.b.next = new Node(3);
list.b.next.next = new Node(20);
System.out.println(\"List a before merge :\");
list.printlist(a);
System.out.println(\"\");
System.out.println(\"List b before merge :\");
list.printlist(b);
// Cal the function to merge the list in increasing order.
sortResult = list.sortedmerge(a, b);
System.out.println(\"\");
System.out.println(\"Merged linked list : \");
list.printlist(sortResult);
}
}.
Homework 05 - Linked Lists (C++)(1) Implement the concepts of a un.pdf
Homework 05 - Linked Lists (C++)
(1) Implement the concepts of a union, and difference as defined noted in Chapter 1 questions 6,
and 8 using a linked list. You should not use arrays, dynamic arrays or the STL vector or STL
list classes.
You will be linking a series of nodes together to implement a single linked list class. Define a
Node as a struct that holds a character (like they did in the book) and a pointer to another Node:
struct Node {
char data;
Node* next;
};
Then define a LinkedList class which has (at a minimum) these member functions:
LinkedList();
~LinkedList();
bool insertAtFront();
bool insertBeforePosition(int index);
bool insertAtBack();
bool deleteAtFront();
bool deleteBeforePosition(int index);
bool deleteAtBack();
Implement each of these functions to provide the correct functionality. These functions return
true if successful in inserting or deleting, otherwise they return false (indicating the operation
was not successful).
Finally, create an overloaded + operator to handle the Union of two linked lists and an
overloaded - operator to handle the Difference of two linked lists.
Because we are dealing with pointers you should have both a LinkedList Constructor and
Destructor. Remember that you do not directly call a Constructor or Destructor Function. The
Destructor is automatically called when the variable loses scope or the program ends.
Remember, that we are dealing with not just one dynamically allocated Node (with the new
operator), but many, so you will have to start at the head of the list and go until the Node points
to nullptr. Then keep deleting the previous Node pointer until there are no Nodes left to delete.
(2) To verify your set class, write a main function that takes two lines of characters from the
input file input.txt and store each line of characters into two separate linked lists. Then using
these two lists, perform the Union and set Difference of the characters from the file, and print the
results to the console.
(3) Please also complete an asymptotic (Big O) analysis of your insertAtFront() and
insertAtBack() member functions. Place this in a file called analysis.txt.
Solution
Here is the complete code for your question. Please do rate the aswer if you are happy with
program. Please read comments in the program. Also pay special attention to comments in + and
- operators. You WILL NEED to comment out if condition in + if you want all elements in both
lists irrespective of if they duplicates or no. For now implemented to have true behaviour of no
duplicates in union.
check the functions by creating some lists and playing around different fucntions like its done in
main and commented out.
As far as complexity:
for insertAtFront() its O(1) since it does not depend on the size of the list. It just replaces the start
of the list. Hence it is O(1) and independent of the size of list.
for insertAtBack(), its O(n) since it depends on the number of elements in the list. It should
traverse till end of list and .
Similar to code ----- public void add(int key, T value) { Link previou.pdf (20)
1#include stdio.h#include stdlib.h#include assert.h .pdf
1#include stdio.h#include stdlib.h#include assert.h .pdf
Template LinkedList;I am using templates to make some linkedLists.pdf
Template LinkedList;I am using templates to make some linkedLists.pdf
I have been tasked to write a code for a Singly Linked list that inc.pdf
I have been tasked to write a code for a Singly Linked list that inc.pdf
please help me in C++Objective Create a singly linked list of num.pdf
please help me in C++Objective Create a singly linked list of num.pdf
Data Structures in C++I am really new to C++, so links are really .pdf
Data Structures in C++I am really new to C++, so links are really .pdf
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Let n = size of the hash table hashTable = array of size n, w.pdf
Let n = size of the hash table hashTable = array of size n, w.pdf
double Skiplist code below#include iostream#include vector.pdf
double Skiplist code below#include iostream#include vector.pdf
Dividing a linked list into two sublists of almost equal sizesa. A.pdf
Dividing a linked list into two sublists of almost equal sizesa. A.pdf
using the single linked list code written in the class or in the lab,.pdf
using the single linked list code written in the class or in the lab,.pdf
Homework 05 - Linked Lists (C++)(1) Implement the concepts of a un.pdf
Homework 05 - Linked Lists (C++)(1) Implement the concepts of a un.pdf
More from annucommunication1
4r = 2 L2r = L2thats none of the above, since you didnt wrote t.pdf
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2
Solution
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2.
41.Ans. Ans,Sarcoplasmic reticulumSarcoplasmic reticulum is a type.pdf
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
volume is cytosol.
the other options are wrong. membrane bound organelles have outer layer plasma membrane, so
the endoplasmic reticulum not far exceeds the area of plasma membrane. not only the nucleus
surrounded by double layer, other organells for example Mitochondria also surrounded by
double layers. and not only the nucleus, mitochondria also have the DNA. and chloroplast also
have the DNA.
43.Ans. True
once loaded vesicles are carry the cargo to different location with in the body, and reaches the
site of secretion. once reached some of the vesicles containing material for constitutive -
Produced in constant amount, releases and immediately fused with Plasma membrane. where as
vesicles called secretory vesicles- these are the vesicles in which molecules destined for
secretion are stored, these vesicles are in regulated pathway means once reached to the site of
secretion these vesicles wait at he membrane until it receives a signal to secrete, once cell
receives the signal these are going secrete the content and material fuse with the plasma
membrane
44.Ans.False
Nuclear localization signal needed to the peptide until it crosses the membrane
45Ans.True
Chaperones are needed for the import of the proteins. these are prevent the premature folding of
the proteins. while import into the mitochondria, the Chaperones present in the inner surface of
the membrane of mitochondria able to bind the protruding ends of polypeptide and prevents them
to diffuse back into the cytosol through the pore and carry them further and into the matrix. while
moving the peptides further into the matrix at every stage the se peptides are bounded with the
Chaperones which prevent backward movement of the peptides. this kind of diffusion called as
biased diffusion and Chaperones acted as Brownian ratchet.
46.Ans.E) All of These happen.
the above all happens when haploid cell mates
Solution
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
vol.
1.c. developers2.e. A and B3.c. Land and Built space4. Money a.pdf
1.c. developers
2.e. A and B
3.c. Land and Built space
4. Money and Stock
5.c. construction and land market
6. expect a reduction in property asset value
7.b. Demand
8. converts physical capital into financial capital
9.D
10.A
11.D
12.C
13.A
14.B
15.A
16. rent go up, prices of real estsate asset go up, new development increases
17.C
18.D
19.D
20. location and size
21.B
22.D
23.D
24.B
25.C
26.B
27.D
28.B
29.A
30.D
31.D
32.B
33.C
34.D
35.C
36.D
37.A
38.E
39.A
40.C
41.C
42.A
43.A
44.D
45.C
Solution
1.c. developers
2.e. A and B
3.c. Land and Built space
4. Money and Stock
5.c. construction and land market
6. expect a reduction in property asset value
7.b. Demand
8. converts physical capital into financial capital
9.D
10.A
11.D
12.C
13.A
14.B
15.A
16. rent go up, prices of real estsate asset go up, new development increases
17.C
18.D
19.D
20. location and size
21.B
22.D
23.D
24.B
25.C
26.B
27.D
28.B
29.A
30.D
31.D
32.B
33.C
34.D
35.C
36.D
37.A
38.E
39.A
40.C
41.C
42.A
43.A
44.D
45.C.
package patienttest;import java.util.Comparator; import java..pdf
package patienttest;
import java.util.Comparator;
import java.util.Date;
public class Patient implements Comparator{
String Name;
Date DOB;
Date Admission;
String Complaint;
int Priority;
public Patient(String name,Date d1,Date ad,String complaint,int p)
{
Name=name;
DOB=d1;
Admission=ad;
Complaint=complaint;
Priority=p;
}
public void setName(String name)
{
Name=name;
}
public void setDOB(Date d1)
{
DOB=d1;
}
public void setAdmission(Date d2)
{
Admission=d2;
}
public void setComplaint(String complaint)
{
Complaint=complaint;
}
public void setPriority(int p)
{
Priority=p;
}
public String setName()
{
return Name;
}
public Date setDOB()
{
return DOB;
}
public Date setAdmission()
{
return Admission;
}
public String setComplaint()
{
return Complaint;
}
public int setPriority()
{
return Priority;
}
@Override
public String toString()
{
String msg=\"\";
msg+=\"Name: \"+Name;
msg+=\"DOB: \"+DOB;
msg+=\"Admission: \"+Admission;
msg+=\"Complaint: \"+Complaint;
msg+=\"Priority: \"+Priority;
return msg;
}
@Override
public int compare(Patient p1, Patient p2) {
int value=1;
if(p1.Priority>p2.Priority)
value=1;
else if(p1.Priority0)
value=1;
else
value=-1;
}
return value;
}
}
Solution
package patienttest;
import java.util.Comparator;
import java.util.Date;
public class Patient implements Comparator{
String Name;
Date DOB;
Date Admission;
String Complaint;
int Priority;
public Patient(String name,Date d1,Date ad,String complaint,int p)
{
Name=name;
DOB=d1;
Admission=ad;
Complaint=complaint;
Priority=p;
}
public void setName(String name)
{
Name=name;
}
public void setDOB(Date d1)
{
DOB=d1;
}
public void setAdmission(Date d2)
{
Admission=d2;
}
public void setComplaint(String complaint)
{
Complaint=complaint;
}
public void setPriority(int p)
{
Priority=p;
}
public String setName()
{
return Name;
}
public Date setDOB()
{
return DOB;
}
public Date setAdmission()
{
return Admission;
}
public String setComplaint()
{
return Complaint;
}
public int setPriority()
{
return Priority;
}
@Override
public String toString()
{
String msg=\"\";
msg+=\"Name: \"+Name;
msg+=\"DOB: \"+DOB;
msg+=\"Admission: \"+Admission;
msg+=\"Complaint: \"+Complaint;
msg+=\"Priority: \"+Priority;
return msg;
}
@Override
public int compare(Patient p1, Patient p2) {
int value=1;
if(p1.Priority>p2.Priority)
value=1;
else if(p1.Priority0)
value=1;
else
value=-1;
}
return value;
}
}.
ZnO + H2O = Zn(OH)2 .pdf
This document describes a chemical reaction where zinc oxide reacts with water to form zinc hydroxide. Zinc oxide and water react to form zinc hydroxide according to the chemical equation ZnO + H2O => Zn(OH)2. The reaction shows zinc oxide dissolving in water to form zinc hydroxide as the product.
will result in the formation of a soluble complex.pdf
will result in the formation of a soluble complex ion formed from reaction with
hydroxide ion
Solution
will result in the formation of a soluble complex ion formed from reaction with
hydroxide ion.
What types of elements are involved in metallic b.pdf
What types of elements are involved in metallic bonding? D. metals only Metallic
bonding is the electromagnetic interaction between delocalized electrons, called conduction
electrons and the metallic nuclei within metals.
Solution
What types of elements are involved in metallic bonding? D. metals only Metallic
bonding is the electromagnetic interaction between delocalized electrons, called conduction
electrons and the metallic nuclei within metals..
The correct statements are When two atoms of app.pdf
The correct statements are When two atoms of approach each other, the two nuclei
repel each other. When two atoms of approach each other, each electron is attracted by the other
nucleus
Solution
The correct statements are When two atoms of approach each other, the two nuclei
repel each other. When two atoms of approach each other, each electron is attracted by the other
nucleus.
Sure can There you go. .pdf
This very short document contains only two sentences that state "Sure can: There you go." without providing any other context or information. It does not have enough content to create a meaningful multi-sentence summary.
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol.pdf
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol oxygen atoms
Solution
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol oxygen atoms.
Na2O is basic NO2 is acidic SO2 is acidic CO2 is .pdf
Na2O is basic NO2 is acidic SO2 is acidic CO2 is acidic CaO is basic
Solution
Na2O is basic NO2 is acidic SO2 is acidic CO2 is acidic CaO is basic.
moles of CO2 = moles of NaHCO3 = 284 =0.0238 .pdf
moles of CO2 = moles of NaHCO3 = 2/84 =0.0238
Solution
moles of CO2 = moles of NaHCO3 = 2/84 =0.0238.
Eukaryotic Chromosome Structure The length of DN.pdf
Eukaryotic Chromosome Structure The length of DNA in the nucleus is far greater
than the size of the compartment in which it is contained. To fit into this compartment the DNA
has to be condensed in some manner. The degree to which DNA is condensed is expressed as its
packing ratio. Packing ratio - the length of DNA divided by the length into which it is packaged
For example, the shortest human chromosome contains 4.6 x 107 bp of DNA (about 10 times the
genome size of E. coli). This is equivalent to 14,000 µm of extended DNA. In its most condensed
state during mitosis, the chromosome is about 2 µm long. This gives a packing ratio of 7000
(14,000/2). To achieve the overall packing ratio, DNA is not packaged directly into final
structure of chromatin. Instead, it contains several hierarchies of organization. The first level of
packing is achieved by the winding of DNA around a protein core to produce a \"bead-like\"
structure called a nucleosome. This gives a packing ratio of about 6. This structure is invariant in
both the euchromatin and heterochromatin of all chromosomes. The second level of packing is
the coiling of beads in a helical structure called the 30 nm fiber that is found in both interphase
chromatin and mitotic chromosomes. This structure increases the packing ratio to about 40. The
final packaging occurs when the fiber is organized in loops, scaffolds and domains that give a
final packing ratio of about 1000 in interphase chromosomes and about 10,000 in mitotic
chromosomes. Eukaryotic chromosomes consist of a DNA-protein complex that is organized in
a compact manner which permits the large amount of DNA to be stored in the nucleus of the cell.
The subunit designation of the chromosome is chromatin. The fundamental unit of chromatin is
the nucleosome. Chromatin - the unit of analysis of the chromosome; chromatin reflects the
general structure of the chromosome but is not unique to any particular chromosome
Nucleosome - simplest packaging structure of DNA that is found in all eukaryotic chromosomes;
DNA is wrapped around an octamer of small basic proteins called histones; 146 bp is wrapped
around the core and the remaining bases link to the next nucleosome; this structure causes
negative supercoiling The nucleosome consists of about 200 bp wrapped around a histone
octamer that contains two copies of histone proteins H2A, H2B, H3 and H4. These are known as
the core histones. Histones are basic proteins that have an affinity for DNA and are the most
abundant proteins associated with DNA. The amino acid sequence of these four histones is
conserved suggesting a similar function for all. The length of DNA that is associated with the
nucleosome unit varies between species. But regardless of the size, two DNA components are
involved. Core DNA is the DNA that is actually associated with the histone octamer. This value
is invariant and is 146 base pairs. The core DNA forms two loops around the octamer, and this
permits two regions that are .
Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf
Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625
here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS
= 0.1682moles mass of FeS = 0.1682*88 = 14.8g
Solution
Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625
here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS
= 0.1682moles mass of FeS = 0.1682*88 = 14.8g.
D. is correct. Solution .pdf
The document states that option D is the correct answer. It provides no other context or details about what question or problem is being addressed.
The various ways to finance a project are 1)Debt financing-One ca.pdf
The various ways to finance a project are:
1)Debt financing-One can take out a loan and finance the project costs
Pros:
Ownership is maintained
Lower interest rate
Lower cost of capital
Cons:
Repayment needed even if business goes bankrupt
High rates for old debt even if macroeconomic conditions change
Collateral needed
2) Equity financing is when project financing provides share in a firm in lieu of financing
Pros- Easily availaible than debt
Not to be repaid in case of bankrupcy
Periodic payment as in case of debt is not required. Dividends can be paid as when the owner
feels so
Cons-
have to share profit
High return required by investor
May lose control of company
Private equity funding is when a large investor invests in company and his share cannot be
publicly tarded:
Pros- Capital is easily availaible and all capital can be raised at once
The private equity investor provides his insights
No return requirement and no cash flow has to be paid out as in case of equity
Cons-
More interference in day to day matters by investor
investor may want to recoup his investment when he feels so but company may still have better
profits
Solution
The various ways to finance a project are:
1)Debt financing-One can take out a loan and finance the project costs
Pros:
Ownership is maintained
Lower interest rate
Lower cost of capital
Cons:
Repayment needed even if business goes bankrupt
High rates for old debt even if macroeconomic conditions change
Collateral needed
2) Equity financing is when project financing provides share in a firm in lieu of financing
Pros- Easily availaible than debt
Not to be repaid in case of bankrupcy
Periodic payment as in case of debt is not required. Dividends can be paid as when the owner
feels so
Cons-
have to share profit
High return required by investor
May lose control of company
Private equity funding is when a large investor invests in company and his share cannot be
publicly tarded:
Pros- Capital is easily availaible and all capital can be raised at once
The private equity investor provides his insights
No return requirement and no cash flow has to be paid out as in case of equity
Cons-
More interference in day to day matters by investor
investor may want to recoup his investment when he feels so but company may still have better
profits.
boiling point 1-propanol propanone note there.pdf
boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen
bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction
between propanone molecules. Thus, bp of 1-propanol is higher than propanone.
Solution
boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen
bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction
between propanone molecules. Thus, bp of 1-propanol is higher than propanone..
The organs of urinary system includesPaired kidneys, paired ureter.pdf
The organs of urinary system includes
Paired kidneys, paired ureters, a urinary bladder and a urethra.
So the answer is option 4.
Solution
The organs of urinary system includes
Paired kidneys, paired ureters, a urinary bladder and a urethra.
So the answer is option 4..
The incorrect statement ise. Glucose is a stereoisomer of ribose..pdf
The incorrect statement is:
e. Glucose is a stereoisomer of ribose.
Glucose is C6H12O6 while ribose is C5H10O5. They are not ismoers.
Solution
The incorrect statement is:
e. Glucose is a stereoisomer of ribose.
Glucose is C6H12O6 while ribose is C5H10O5. They are not ismoers..
The answer is a. The intermolecular forces are much greater in wate.pdf
The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide
and hydrogen selenide.
Boiling point increases with strength of intermolecular forces as these intermolecular bonds need
to be broken for the molecules to escape into the gas phase. Water has strong intermolecular
hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much
higher boiling point.
Solution
The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide
and hydrogen selenide.
Boiling point increases with strength of intermolecular forces as these intermolecular bonds need
to be broken for the molecules to escape into the gas phase. Water has strong intermolecular
hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much
higher boiling point..
More from annucommunication1 (20)
4r = 2 L2r = L2thats none of the above, since you didnt wrote t.pdf
4r = 2 L2r = L2thats none of the above, since you didnt wrote t.pdf
41.Ans. Ans,Sarcoplasmic reticulumSarcoplasmic reticulum is a type.pdf
41.Ans. Ans,Sarcoplasmic reticulumSarcoplasmic reticulum is a type.pdf
1.c. developers2.e. A and B3.c. Land and Built space4. Money a.pdf
1.c. developers2.e. A and B3.c. Land and Built space4. Money a.pdf
package patienttest;import java.util.Comparator; import java..pdf
package patienttest;import java.util.Comparator; import java..pdf
will result in the formation of a soluble complex.pdf
will result in the formation of a soluble complex.pdf
What types of elements are involved in metallic b.pdf
What types of elements are involved in metallic b.pdf
The correct statements are When two atoms of app.pdf
The correct statements are When two atoms of app.pdf
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol.pdf
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol.pdf
Na2O is basic NO2 is acidic SO2 is acidic CO2 is .pdf
Na2O is basic NO2 is acidic SO2 is acidic CO2 is .pdf
Eukaryotic Chromosome Structure The length of DN.pdf
Eukaryotic Chromosome Structure The length of DN.pdf
Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf
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