Create an ArrayList only containing Strings. Print using enhanced for loop

Question

So here is my program:

Create an ArrayList that will only contain strings Add the following to the list in order

• Mary
• John
• Mahendra
• Sara
• Jose
• Judy

Print the list using the enhanced for loop Insert Harry in front of Mahendra and after John Then Remove position 4 from the list

Here's what I've written:

import java.util.ArrayList;
import java.util.Scanner;

public class Name {
    public static void main(String[] args) {
        ArrayList<String> names = new ArrayList<String>();
        Scanner input = new Scanner(System.in);
        names.add(input.nextLine());
        names.add(input.nextLine());
        names.add(input.nextLine());
        names.add(input.nextLine());
        names.add(input.nextLine());
        names.add(input.nextLine());
        names.add(input.nextLine());

        for (String n : names) {
        System.out.println(n);
        }
    }
}

I guess I'm having problems with adding and removing. I believe everything else should be fine though.

Solution 1

You may want to use below methods to insert and remove:

void    add(int index, E element)
 E      remove(int index)

e.g. Mahendra is at index 2(index starts from 0), then to add Harry in front of Mahendra, just do as below:

  names.add(2, "Harry"); //This will push Mahendra at index 3

To remove crrent index 4,

  names.remove(4);

To remove previous index 4, which has become index 5 now,

  names.remove(5);

Solution 2

indexOf() will let you find position of a given entry, add(index, object) will let you insert at an index.

public static void main(String[] args) {
    List<String> names = new ArrayList<String>();

    names.add("Mary");
    names.add("John");
    names.add("Mahendra");
    names.add("Sara");
    names.add("Jose");
    names.add("Judy");

    names.add(names.indexOf("Mahendra"), "Harry");

    for (String name : names) {
        System.out.println(name);
    }
}

Comments

[Yogendra Singh]

@user1793131: Please don't forget to accept the answer.