Prompt for the user to enter their name. Print out their name forwards, backwards, vertically-forwards and vertically-backwards. You will need to declare a String variable, use Scanner to get input of a String, call a String method to isolate one letter and create a loop that traverses the name forwards and backwards. Sample Output: Enter your name: Florene Florene enerolF F l o r e n e e n e r o l F HINTS: Scanner input = new Scanner(System.in); String myName = input.nextLine(); // input is a Scanner object int len = myName.length() // len is the number of characters System.out.println (myName.substring(len-1, len)); // prints last letter I need help with coding the name "Pooja" vertically-forwards and vertically-backwards in java. If you could show me what codes to type in, that would be great. Thank you!

Prompt for the user to enter their name. Print out their name forwards, backwards, vertically-forwards and vertically-backwards. You will need to declare a String variable, use Scanner to get input of a String, call a String method to isolate one letter and create a loop that traverses the name forwards and backwards. Sample Output: Enter your name: Florene Florene enerolF F l o r e n e e n e r o l F HINTS: Scanner input = new Scanner(System.in); String myName = input.nextLine(); // input is a Scanner object int len = myName.length() // len is the number of characters System.out.println (myName.substring(len-1, len)); // prints last letter I need help with coding the name "Pooja" vertically-forwards and vertically-backwards in java. If you could show me what codes to type in, that would be great. Thank you!

Microsoft Visual C#

7th Edition

ISBN:9781337102100

Author:Joyce, Farrell.

Publisher:Joyce, Farrell.

Chapter5: Looping

Section: Chapter Questions

Problem 15E

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This method will return a string that when printed out will display a triangle. Each row of the triangle will have the next letter from the input argument as the last letter in the row. If the next letter is the noPrint char, * will print out instead of letters for the row. Can’t use System.out.println to print the calculated result in the method. You return it
This method will return a string that when printed out will display a triangle. Each row of the triangle will have the next letter from the input argument as the last letter in the row. If the next letter is the noPrint char, * will print out instead of letters for the row. Can’t cannot use System.out.println to print the calculated result in the method. You return it
This is the question I am stuck on - Write an application that accepts any number of String values from a user until they enter zzz or have entered 15 strings, and display them in ascending order. This is the code I have started to try but I am honestly just confusing myself more - import java.util.*; public class StringSort2 { public static void main(String[] args) { // Write your code here Scanner input = new Scanner(System.in); String values; int x; System.out.println("Please enter your values: "); values = input.nextLine(); for(x=15; x < values.length; ++x) System.out.println(values[x] + " "); } }
This is the question - Write an application that accepts any number of String values from a user until they enter zzz or have entered 15 strings, and display them in ascending order. This is the code I have but it is wrong becuase I didn't get a single check right - import java.util.*; public class StringSort2 { public static void main(String[] args) { // Write your code here Scanner sc = new Scanner(System.in); String s = ""; System.out.print(""); ///loop works until string input is zzz or string length is greater than 15 while (true){ String input = sc.nextLine(); if(input.equalsIgnoreCase("zzz") || input.length()>15) break; s=input; } } }
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Write a program that prompts the user to enter the number of students,the students’ names, and their scores and prints student names in decreasingorder of their scores. Assume the name is a string without spaces, use the Scanner’snext() method to read a name.
This method will return a string that when printed out will display a triangle. Each row of the triangle will have the next letter from the input argument as the last letter in the row. If the next letter is the noPrint char, * will print out instead of letters for the row. Can’t use System.out.println.
Print numbers 0, 1, 2, ..., userNum as shown, with each number indented by that number of spaces. For each printed line, print the leading spaces, then the number, and then a newline. Hint: Use i and j as loop variables (initialize i and j explicitly). Note: Avoid any other spaces like spaces after the printed number. import java.util.Scanner;public class NestedLoop {public static void main (String [] args) {Scanner scnr = new Scanner(System.in);int userNum;int i;int j; userNum = scnr.nextInt();
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The method will return a string that when printed out will display a square or Bad input. If the input argument has a least one noPrint char in it, the string Bad input is returned. Other return a square, where each row reveals just the next letter of input and the remaining letters are *. Not using System.out.println. Must be this output provided in the image
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Implement your own indexOf(String s) and lastIndexOf(String s) methods without using the built-in ones available in Java String class. Complete the following program://course: CSC190//project: Lab 12//date: (today’s date)//author: (your name)//purpose:import java.util.Scanner;class MyString {String s; //default constructor which sets s to a null string //your code here //additional constructor which sets s to the string passed from the caller //your code here //get and set methods for s //your code here //return the index within s of the first occurrence of str. //return -1 when str is not found on s. //if s = "aabababb", s = "aba", return 1 //if s = "aabababb", s = "abaa", return -1 int indexOf(String str) { //your code here} //return the index within s of the last occurrence of str. //return -1 when s is not found on str. //if s = "aabababb", s = "aba", return 3 //if s = "aabababb", s = "abaa", return -1 int lastIndexOf(String s) { //your…