Iterative Cartesian Product in Java

Question

I want to compute the cartesian product of an arbitrary number of nonempty sets in Java.

I\'ve wrote that iterative code...

public s         


        

Answer 1

Here is a lazy iterator approach that uses a function to produce an appropriate output type.

  public static  Iterable cartesianProduct(
      final Function fn, Object[]... options) {
    final Object[][] opts = new Object[options.length][];
    for (int i = opts.length; --i >= 0;) {
      // NPE on null input collections, and handle the empty output case here
      // since the iterator code below assumes that it is not exhausted the
      // first time through fetch.
      if (options[i].length == 0) { return Collections.emptySet(); }
      opts[i] = options[i].clone();
    }
    return new Iterable() {
      public Iterator iterator() {
        return new Iterator() {
          final int[] pos = new int[opts.length];
          boolean hasPending;
          T pending;
          boolean exhausted;

          public boolean hasNext() {
            fetch();
            return hasPending;
          }

          public T next() {
            fetch();
            if (!hasPending) { throw new NoSuchElementException(); }
            T out = pending;
            pending = null;  // release for GC
            hasPending = false;
            return out;
          }

          public void remove() { throw new UnsupportedOperationException(); }

          private void fetch() {
            if (hasPending || exhausted) { return; }
            // Produce a result.
            int n = pos.length;
            Object[] args = new Object[n];
            for (int j = n; --j >= 0;) { args[j] = opts[j][pos[j]]; }
            pending = fn.apply(args);
            hasPending = true;
            // Increment to next.
            for (int i = n; --i >= 0;) {
              if (++pos[i] < opts[i].length) {
                for (int j = n; --j > i;) { pos[j] = 0; }
                return;
              }
            }
            exhausted = true;
          }
        };
      }
    };
  }