Пример #1
0
    /**
     * Given an array of slices (potentially overlapping and in any order) and return an equivalent
     * array of non-overlapping slices in clustering order.
     *
     * @param slices an array of slices. This may be modified by this method.
     * @return the smallest possible array of non-overlapping slices in clustering order. If the
     *     original slices are already non-overlapping and in comparator order, this may or may not
     *     return the provided slices directly.
     */
    private List<Slice> normalize(List<Slice> slices) {
      if (slices.size() <= 1) return slices;

      Collections.sort(
          slices,
          new Comparator<Slice>() {
            @Override
            public int compare(Slice s1, Slice s2) {
              int c = comparator.compare(s1.start(), s2.start());
              if (c != 0) return c;

              return comparator.compare(s1.end(), s2.end());
            }
          });

      List<Slice> slicesCopy = new ArrayList<>(slices.size());

      Slice last = slices.get(0);

      for (int i = 1; i < slices.size(); i++) {
        Slice s2 = slices.get(i);

        boolean includesStart = last.includes(comparator, s2.start());
        boolean includesFinish = last.includes(comparator, s2.end());

        if (includesStart && includesFinish) continue;

        if (!includesStart && !includesFinish) {
          slicesCopy.add(last);
          last = s2;
          continue;
        }

        if (includesStart) {
          last = Slice.make(last.start(), s2.end());
          continue;
        }

        assert !includesFinish;
      }

      slicesCopy.add(last);
      return slicesCopy;
    }
Пример #2
0
 public Builder add(Slice.Bound start, Slice.Bound end) {
   return add(Slice.make(start, end));
 }