Re: Algorithm for performing a rollup

From: (Stefan Ram)
19 Mar 2007 00:26:04 GMT
<> (Stefan Ram) writes:

Between a call of this and the next call there
must be exactly one call of "count".

  This burden placed on the client has been removed in the
  following version, where the client is free to call ?value?
  and ?count? an arbitrary number of times and in an arbitrary
  order after each successful ?advance?.

class Source
implements java.util.Iterator<java.lang.String>
{ private final java.lang.String source[] =
  { "A", "A", "A", "B", "B", "C", "D", "D" };
  private int position;
  public Source()
  { this.position = 0; }
  public java.lang.String next(){ return source[ position++ ]; }
  public boolean hasNext(){ return position < source.length; }
  public void remove(){ throw new java.lang.UnsupportedOperationException(); }}

/** For an Iterator<E> object delivering components, allow to count how many
components appear in a sequence of consecutive equal components.
@param E the type of the components */

class Counter<E>
  /** Initialize a new Counter object for a source.
  @param source an iterator delivering each component of the source in turn */

  public Counter( final java.util.Iterator<E> source )
  { this.first = null; = null;
    this.count = -1;
    this.source = source; }

  /** Advance the Counter object to the beginning of the next
  sequence of consecutive equal components.
  A true result indicates success. */

  public boolean advance()
  { this.first = null;
    this.count = -1;
    final boolean advanced;
    if( next != null )
    { first = next; next = null; advanced = true; }
    else if( source.hasNext() )
    { first =; advanced = true; }
    else advanced = false;
    if( advanced )doCount();
    return advanced; }

  /** The current value of the component, valid after a successful
  advance operation */

  public E value()
  { return first; }

  /** The current value of the counter, valid after a successful
  advance operation */

  public int count()
  { return count; }

  /** count the number of consecutive equal source components */
  private void doCount()
  { count = 1; while( this.isExtendable() )++count; }

  /** Is the next component equal to the first?
  side-effect: advance the (perceived) position to the next component if it is
  equal to the first */
  private boolean isExtendable()
  { final boolean result;
    if( !source.hasNext() ){ next = null; result = false; }
    else result = first.equals( next = );
    return result; }

  /** The first component in a sequence of consecutive
  equal source components */
  private E first;

  /** The next component as just read from the source */
  private E next;

  /** The component source for this Counter */
  private final java.util.Iterator<E> source;

  /** The last count */
  private int count = -1; }

public class Main
  public static void main( final java.lang.String[] args )
  { Source source = new Source();
    Counter<java.lang.String> counter = new Counter<java.lang.String>( source );
    while( counter.advance() )
    { java.lang.System.out.println
      ( "\"" + counter.value() + "\", " + counter.count() ); }}}

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