Java Programming Problems: Working With Lists

Write a method void interchange (List<E> l) which interchanges the current element in the list and the one following it. (First make it work in the normal case; then, if make sure it handles special cases, such as: the list is empty; the list has only one element; current is at the end of the list)

Write a method void reverse (List<E> l) which reversed the order of items stored in the list. Again, make sure it works for any special cases you can think of.

Write a method List mergeLists (List<E> l1, List<E> l2)which takes two sorted Lists as input, producing a new List containing all the elements of both lists, also sorted. The new list is the return value for the method.

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/** Array-based list implementation */
class AList<E> implements List<E> {
private static final int defaultSize = 10; // Default size
private int maxSize;       // Maximum size of list
private int listSize;       // Current # of list items
private int curr;          // Position of current element
private E[] listArray;      // Array holding list elements

/** Constructors */
/** Create a list with the default capacity. */
AList() { this(defaultSize); }
/** Create a new list object.
      @param size Max # of elements list can contain. */
@SuppressWarnings("unchecked") // Generic array allocation
AList(int size) {
    maxSize = size;
    listSize = curr = 0;
    listArray = (E[])new Object[size]; // Create listArray
}

public void clear()         // Reinitialize the list
    { listSize = curr = 0; } // Simply reinitialize values

/** Insert "it" at current position */
public void insert(E it) {
    assert listSize < maxSize : "List capacity exceeded";
    for (int i=listSize; i>curr; i--) // Shift elements up
      listArray[i] = listArray[i-1]; //   to make room
    listArray[curr] = it;
    listSize++;               // Increment list size
}...