Final review on data structures

FinalReview.java

import java.util.*;

// Let me know if you find any mistakes!!

public class FinalReview {

    public static void main(String[] args) {
        System.out.println("Question 1:");
        List<String> list = new ArrayList<>();
        list.add("1");
        list.add("2");
        list.add("3");
        System.out.println("Forward: " + list.toString());
        System.out.println("Reverse: " + reverse(list).toString());
    }

    public static <T> List<T> reverse(List<T> list) {
        Stack<T> stack = new Stack<>();
        for(int i = 0 ; i < list.size() ; ++i) {
            stack.push(list.get(i));
        }
        for(int i = 0 ; i < list.size() ; ++i) {
            list.set(i, stack.pop());
        }
        return list;
    }

    // Question 2a:
    //   42 returned
    //   14 returned
    //   2 returned
    //   19 returned
    //   5 returned
    //   Stack contains: 5 14

    // Question 2b:
    //   4 returned
    //   8 returned
    //   98 returned
    //   15 returned
    //   Queue contains: 15, 5, 2

    // Question 2c:
    //   An array is a good choice if you want your data structure to be bounded.
    //   A linked list is a more natural choice if you do not want bounds

    // Question 2d:
    //   4 * 3 * 2 * 1 = 24

    // Question 2e:
    //   Same as they went in. 'D', 'C', 'B', 'A'

    // Question 2f:
    //  Assuming it doesn't cause a null pointer exception, then it serves to delete a node out of the list.
    //  If you have n1 -> n2 -> n3, n1.next.next = n3
    //  by setting a pointer to n3 to n1.next, n2 is effectively deleted

    // Question 2g:
    //  The node 't' gets inserted into the linked list. (draw it on paper and its relatively clear)

    // Question 2h:
    //  An array, since you can reference any element in it by its offset from beginning.
    //  Moreover, if you want the 4th element, array[3] will fetch it.
}