It is common for people to name directories as dir1, dir2, and so on. When there are ten or more directories, the operating system displays them in dictionary order, as dir1, dir10, dir11, dir12, dir2, dir3, and so on. That is irritating, and it is easy to fix. Provide a comparator that compares strings that end in digit sequences in a way that makes sense to a human. First compare the part before the digit as strings, and then compare the numeric values of the digits.
Your program should work with the provided test program
DirectorySortDemo.java.
Call the class you write DirectoryComparator.java.
Submit the two files in your submission.
DirectoryComparator.java
DirectorySortDemo.java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.collections;
public class DirectorySortDemo
{
public static void main (String [] args)
{
String dirNames ("dir12", "dir5", "dir9", "dirl", "dir4",
"lab10", "1ab2", "lab7", "lab17", "lab8",
"quiz8", "quiz10", "quiz11", "quiz12",
"dirll", "dir8", "dir7", "dir15", "dir3");
ArrayList directories = new
ArrayList<> (Arrays.asList (dirNames));
System.out.println ("Unsorted List:");
System.out.println (directories);
Collections.sort (directories, new DirectoryComparator ());
System.out.println ():
System.out-println ("Sorted List: ");
System.out.-println (directories);

Answer:

removeDuplicates() function:-

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       //make ptr1 equals to head.

       ptr1 = head;

        //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

Explanation:

Complete Code:-

//Create Linked List Class.

class LinkedList {

       //Create head pointer.

       static ListNode head;

       //define structure of ListNode.

       //it has int val(data) and pointer to ListNode i.e, next.

   static class ListNode {

       int val;

       ListNode next;

       //constructor to  create and initialize a node.

       ListNode(int d) {

               val = d;

           next = null;

       }

   }

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       //make ptr1 equals to head.

       ptr1 = head;

       //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

   //display() function prints linked list.

   void display(ListNode node)

   {

       //run while loop till last node.

       while (node != null)

       {

               //print node value of current node.

           System.out.print(node.val + " ");

           //move node pointer by one node.

           node = node.next;

       }

   }

   public static void main(String[] args) {

       //Create object of Linked List class.

       LinkedList list = new LinkedList();

       //first we create nodes and connect them to form a linked list.

       //Create Linked List 1-> 2-> 3-> 2-> 4-> 2-> 5-> 2.

       //Create a Node having node data = 1 and assign head pointer to it.

       //As head is listNode of static type. so, we call head pointer using class Name instead of object name.

       LinkedList.head = new ListNode(1);

       //Create a Node having node data = 2 and assign head.next to it.

       LinkedList.head.next = new ListNode(2);

       LinkedList.head.next.next = new ListNode(3);

       LinkedList.head.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next = new ListNode(4);

       LinkedList.head.next.next.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next.next.next = new ListNode(5);

       LinkedList.head.next.next.next.next.next.next.next = new ListNode(2);

       //display linked list before Removing duplicates.

       System.out.println("Linked List before removing duplicates : ");

       list.display(head);

       //call removeDuplicates() function to remove duplicates from linked list.

       list.removeDuplicates();

       System.out.println("")

       //display linked list after Removing duplicates.

       System.out.println("Linked List after removing duplicates :  ");

       list.display(head);

   }

}

Output:-

Answer:

third one

Explanation:

To write a letter to your grandma using Word Online, you should choose the "New Blank Document" layout.

Microsoft Word is a word processing application. It is part of the Microsoft Office productivity software suite, along with Excel, PowerPoint, Access, and Outlook.

To write a letter to your grandmother in Word Online, select the "New Blank Document" layout. This will open a blank page on which you can begin typing your letter.

The "APA Style Paper" layout is intended for academic papers and includes formatting guidelines and section headings that a personal letter may not require.

In Word Online, "General Notes" is not a document layout option.

The "Table of Contents" feature generates a table of contents based on the headings in your document, but it is not a document layout option.

Thus, the answer is "New Blank Document".

For more details regarding Microsoft word, visit:

https://brainly.com/question/26695071

#SPJ2

there are ten letters in the word, so for the first position we can choose out of 10 option.

no matter wich specific option we choose, we will have 9 letters left for the 2nd letter.

and 8 for the third place.

...

when we will have used up 9 of 10 letters, there will be only one option for the last place.

the number of possible decision-paths is therefore given by this expression:

10*9*8*7*6*5*4*3*2*1

wich is 3,628,800

there are quite a few ways to rearrange 10 letters

The sorting method which commands entries in a going to head predicated on each character without regard to space as well as punctuation, therefore there are 3628800 ways.

So, as [tex]\bold{ 10\times 9 \times 8.... \times 3 \times 2 \times 1 = 362,8800}[/tex] ways, we have such a total number of ways.

Learn more:

brainly.com/question/17276660

Answer:

Here the code is given as follows,

Explanation:

class Example:

   def _init_(self, val):

       self.val = val

   def _gt_(self, other):

       return self.val > other.val

   def _sub_(self,other):

       return abs(len(self.val) - len(other.val))

def main():

   obj1 = Example('this is a string')

   obj2 = Example('this is another one')

   print(obj1 > obj2)

   print(obj1 - obj2)

main()

Output:-

Answer:

confidentiality

Explanation:

Alice sends a message to Bob in a manner such that Bob is the only person who can tell what the real message is. Which security concept is this an example of

This type of security concept is an example of confidentiality. Thus, option B is correct.

Security is the capacity to withstand possible damage from other individuals by placing restrictions on their right to act freely. The term security is the absence of threat or a sense of safety.

Any uncovered danger or weakness in any technology that could be exploited by criminals to compromise equipment or knowledge is a security threat.

It has to do with Alice’s ability to decide on their own how if, and why people exploit individual data if it is being sent to bob. To ensure human rights, welfare, and consciousness, solitude must be protected. People are able to independently create their own personalities.

Therefore, option B is the correct option.

Learn more about security, here:

https://brainly.com/question/15278726

#SPJ2

The question is incomplete, the complete question is:

a. message integrity

b. confidentiality

c. availability

d. non-repudiation

e. authentication

Answer:

updating the computer and having an anti virus on your computer

Answer:

He belongs to the nation

Using the concepts of public and private domains, it means that the photograph of the president can be used by anyone without asking for permission.

---------------

Public and private domains:

A similar question is given at https://brainly.com/question/12202794

Answer:

1. Wrong as per my knowledge, SQLite may be an electronic database software that stores data in relations(tables).

2. Wrong. Content provider interface controls the flow of knowledge from your android application to an external data storage location. Data storage is often in any database or maybe it'll be possible to store the info in an SQLite database also.

3.Right. SQLite uses one file to store data. It also creates a rollback file to backup the stored data/logs the transactions.

4. Wrong. The queries are almost like a traditional electronic database system, the queries are going to be answered as per the conditions given and should return a group of answers or one answer depending upon the info and sort of query used.

5. Right. A cursor helps to point to one entry within the database table. this may be very helpful in manipulating the present row of the query easily.