* first specified. That is, if multiple elements are {@linkplain Object#equals equal}, all except
   * the first are ignored.
   */
  public static <E> ImmutableSet<E> of(E e1, E e2) {
    return construct(2, e1, e2);
  }

  /**
   * Returns an immutable set containing the given elements, minus duplicates, in the order each was

   * the first are ignored.
   */
  public static <E> ImmutableSet<E> of(E e1, E e2, E e3, E e4, E e5) {
    return new RegularSetBuilderImpl<E>(5).add(e1).add(e2).add(e3).add(e4).add(e5).review().build();
  }

  /**
   * Returns an immutable set containing the given elements, minus duplicates, in the order each was

   * as equal, the first is returned.
   *
   * <p>The recommended solution for finding the {@code maximum} of some values depends on the type
   * of your data and the number of elements you have. Read more in the Guava User Guide article on
   * <a href="https://github.com/google/guava/wiki/CollectionUtilitiesExplained#comparators">{@code
   * Comparators}</a>.
   *

    public void normalNormal(String first, Integer second) {
      reactToNullParameters(first, second);
    }

    /** Two-arg method with the second param Nullable. */
    @SuppressWarnings("GoodTime") // false positive; b/122617528
    public void normalNullable(String first, @Nullable Integer second) {
      reactToNullParameters(first, second);
    }

   * will remember the first duplicate key we encountered. All later calls to buildOrThrow() can
   * mention that key with its values. Further duplicates might be added in the meantime but since
   * builders only ever accumulate entries it will always be valid to throw from buildOrThrow() with
   * the first duplicate.
   */

     *
     * @param first the first entry of the table
     * @param entry the entry being removed from the table
     * @return the new first entry for the table
     */
    @GuardedBy("this")
    @CheckForNull
    E removeFromChain(E first, E entry) {
      int newCount = count;
      E newFirst = entry.getNext();
      for (E e = first; e != entry; e = e.getNext()) {

  }

  /** Returns {@code true} if the second list is a permutation of the first. */
  private static boolean isPermutation(List<?> first, List<?> second) {
    if (first.size() != second.size()) {
      return false;
    }
    ObjectCountHashMap<?> firstCounts = counts(first);
    ObjectCountHashMap<?> secondCounts = counts(second);
    if (first.size() != second.size()) {
      return false;
    }

   * Number value = optionalInt.or(0.5); // error
   *
   * FluentIterable<? extends Number> numbers = getSomeNumbers();
   * Optional<? extends Number> first = numbers.first();
   * Number value = first.or(0.5); // error
   * }</pre>
   *
   * <p>As a workaround, it is always safe to cast an {@code Optional<? extends T>} to {@code

  <T2 extends T> Ordering<Entry<T2, ?>> onKeys() {
    return onResultOf(Maps.<T2>keyFunction());
  }

  /**
   * Returns an ordering which first uses the ordering {@code this}, but which in the event of a
   * "tie", then delegates to {@code secondaryComparator}. For example, to sort a bug list first by
   * status and second by priority, you might use {@code byStatus.compound(byPriority)}. For a

    if (!elements.hasNext()) {
      return of();
    }
    E first = elements.next();
    if (!elements.hasNext()) {
      // TODO: Remove "ImmutableSet.<E>" when eclipse bug is fixed.
      return ImmutableSet.<E>of(first);
    }

    Set<E> delegate = Sets.newLinkedHashSet();
    delegate.add(checkNotNull(first));
    do {
      delegate.add(checkNotNull(elements.next()));