Write a program that reads a list of integers and outputs those integers in reverse. The input begins with an integer indicating the number of integers that follow. For coding simplicity, follow each output integer by a comma, including the last one. Assume that the list will always contain fewer than 20 integers. Ex: If the input is: 5 2 4 6 8 10 the output is: 10,8,6,4,2, To achieve the above, first read the integers into an array. Then output the array in reverse.
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Write a program that reads a list of integers and outputs those integers in reverse. The input begins with an integer indicating the number of integers that follow. For coding simplicity, follow each output integer by a comma, including the last one. Assume that the list will always contain fewer than 20 integers.
Ex: If the input is:
5 2 4 6 8 10
the output is:
10,8,6,4,2,
To achieve the above, first read the integers into an array. Then output the array in reverse.
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- 2 Write a java program. call it sortWords. Read from the command line a list of words, any number, from 1 to as many as the user wants. Store the words (strings) in an arraylist. Sort the list in alphabetical order. Print the list with a comma and space between each word. Words should be output in the SAME case as the user gives them. For an A make the sort be case insensitive AND remove the , from after the last word. example: java sortWords book car tree apple house --> apple, book, car, house, tree (The already existing answer on Chegg failed the codePost tests)import java.util.ArrayList;import java.util.Arrays;public class PS{/*** Write the method named mesh.** Start with two ArrayLists of String, A and B, each with* its elements in alphabetical order and without any duplicates.* Return a new list containing the first N elements from the two* lists. The result list should be in alphabetical order and without* duplicates. A and B will both have a size which is N or more.* Your solution should make a single pass over A and B, taking* advantage of the fact that they are in alphabetical order,* copying elements directly to the new list.** Remember, to see if one String is "greater than" or "less than"* another, you need to use the compareTo() method, not the < or >* operators.** Examples:* mesh(["a","c","z"], ["b","f","z"], 3) returns ["a","b","c"]* mesh(["a","c","z"], ["c","f","z"], 3) returns ["a","c","f"]* mesh(["f","g","z"], ["c","f","g"], 3) returns ["c","f","g"]** @param a an ArrayList of String in alphabetical order.* @param b an…import java.util.ArrayList;import java.util.Arrays; public class PS07A{/*** Write the method named mesh.* * Start with two ArrayLists of String, A and B, each with * its elements in alphabetical order and without any duplicates. * Return a new list containing the first N elements from the two * lists. The result list should be in alphabetical order and without * duplicates. A and B will both have a size which is N or more. * Your solution should make a single pass over A and B, taking * advantage of the fact that they are in alphabetical order, * copying elements directly to the new list.* * Remember, to see if one String is "greater than" or "less than" * another, you need to use the compareTo() method, not the < or > * operators. * * Examples:* mesh(["a","c","z"], ["b","f","z"], 3) returns ["a","b","c"]* mesh(["a","c","z"], ["c","f","z"], 3) returns ["a","c","f"]* mesh(["f","g","z"], ["c","f","g"], 3) returns ["c","f","g"]* * @param a an ArrayList of String in alphabetical…
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- Consider the interface below:import java.util.NoSuchElementException; public interface MyList<E> { public int getSize(); public void insert(E data) throws ListOverflowException; public E getElement(E data) throws NoSuchElementException; public boolean delete(E data); // returns false if the data is not deleted in the list public boolean search(E data); // returns index of data, else -1 is return}a. Use an array object to store the elements of the list object. The program should be able to show its size, get element, delete, and search from the array. b. This array object will have a fixed size of five (5) and that an attempt to insert an element into the array when full will result to a ListOverflowException.Consider the interface below:import java.util.NoSuchElementException; public interface MyList<E> { public int getSize(); public void insert(E data) throws ListOverflowException; public E getElement(E data) throws NoSuchElementException; public boolean delete(E data); // returns false if the data is not deleted in the list public boolean search(E data); // returns index of data, else -1 is return}a. Use an array object to store the elements of the list object. The program should be able to show its size, get element, delete, and search from the array. b. If the array is full and an element is to be inserted, a new array object with a size that is twice than the “old” array will be created. All the elements of the “old” array will be copied to the new array before inserting the new element.public class CustomLinkedList { public static int findMax(IntNode headObj) { /* Type your code here */ } public static void main(String[] args) { IntNode headObj; IntNode currObj; IntNode lastObj; int i; int max; // Create head node headObj = new IntNode(-1); lastObj = headObj; // Add nodes to the list for (i = 0; i < 20; ++i) { currObj = new IntNode(i); lastObj.insertAfter(currObj); lastObj = currObj; } max = findMax(headObj); System.out.println(max); }}
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