مرتب‌سازی روان
پیاده‌سازی در جاوا

در این کد دو متغیر lo وhi به عنوان ۲ کران بالا و پایین استفاده شده است
// by keeping these constants, we can avoid the tiresome business
  // of keeping track of Dijkstra's b and c. Instead of keeping
  // b and c, I will keep an index into this array.

  static final int LP[] = { 1, 1, 3, 5, 9, 15, 25, 41, 67, 109,
      177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891,
      35421, 57313, 92735, 150049, 242785, 392835, 635621, 1028457,
      1664079, 2692537, 4356617, 7049155, 11405773, 18454929, 29860703,
      48315633, 78176337, 126491971, 204668309, 331160281, 535828591,
      866988873 // the next number is> 31 bits.
  };

  public static <C extends Comparable<? super C>> void sort(C[] m,
      int lo, int hi) {
    int head = lo; // the offset of the first element of the prefix into m

    // These variables need a little explaining. If our string of heaps
    // is of length 38, then the heaps will be of size 25+9+3+1, which are
    // Leonardo numbers 6, 4, 2, 1. 
    // Turning this int a binary number, we get b01010110 = x56. We represent
    // this number as a pair of numbers by right-shifting all the zeros and 
    // storing the mantissa and exponent as "p" and "pshift".
    // This is handy, because the exponent is the index into L[] giving the
    // size of the rightmost heap, and because we can instantly find out if
    // the rightmost two heaps are consecutive leonardo numbers by checking
    // (p&3)==3

    int p = 1; // the bitmap of the current standard concatenation>> pshift
    int pshift = 1;

    while (head <hi) {
      if ((p & 3) == 3) {
        // Add 1 by merging the first two blocks into a larger one.
        // The next Leonardo num is one bigger.
        sift(m, pshift, head);
        p>>>= 2;
        pshift += 2;
      } else {
        // adding a new block of length 1
        if (LP[pshift - 1]>= hi - head) {
          // this block is its final size.
          trinkle(m, p, pshift, head, false);
        } else {
          // this block will get merged. Just make it trusty.
          sift(m, pshift, head);
        }

        if (pshift == 1) {
          // LP[1] is being used, so we add use LP[0]
          p <<= 1;
          pshift--;
        } else {
          // shift out to position 1, add LP[1]
          p <<= (pshift - 1);
          pshift = 1;
        }
      }
      p |= 1;
      head++;
    }

    trinkle(m, p, pshift, head, false);

    while (pshift != 1 || p != 1) {
      if (pshift <= 1) {
        // block of length 1. No fiddling needed
        int trail = Integer.numberOfTrailingZeros(p & ~1);
        p>>>= trail;
        pshift += trail;
      } else {
        p <<= 2;
        p ^= 7;
        pshift -= 2;

        // ok. This block gets broken into three bits. The rightmost
        // bit is a block of length 1. The left hand part is split into
        // two, a block of length LP[pshift+1] and one of LP[pshift].
        // Both these two are appropriately heapified, but the root
        // nodes are not nessesarily in order. We therefore semitrinkle
        // both
        // of them

        trinkle(m, p>>> 1, pshift + 1, head - LP[pshift] - 1, true);
        trinkle(m, p, pshift, head - 1, true);
      }

      head--;
    }
  }

  private static <C extends Comparable<? super C>> void sift(C[] m, int pshift,
      int head) {
    // we do not use Floyd's improvements to the heapsort sift, because we
    // are not doing what heapsort does - always moving nodes from near
    // the bottom of the tree to the root.

    C val = m[head];

    while (pshift> 1) {
      int rt = head - 1;
      int lf = head - 1 - LP[pshift - 2];

      if (val.compareTo(m[lf])>= 0 && val.compareTo(m[rt])>= 0)
        break;
      if (m[lf].compareTo(m[rt])>= 0) {
        m[head] = m[lf];
        head = lf;
        pshift -= 1;
      } else {
        m[head] = m[rt];
        head = rt;
        pshift -= 2;
      }
    }

    m[head] = val;
  }

  private static <C extends Comparable<? super C>> void trinkle(C[] m, int p,
      int pshift, int head, boolean isTrusty) {

    C val = m[head];

    while (p != 1) {
      int stepson = head - LP[pshift];

      if (m[stepson].compareTo(val) <= 0)
        break; // current node is greater than head. Sift.

      // no need to check this if we know the current node is trusty,
      // because we just checked the head (which is val, in the first
      // iteration)
      if (!isTrusty && pshift> 1) {
        int rt = head - 1;
        int lf = head - 1 - LP[pshift - 2];
        if (m[rt].compareTo(m[stepson])>= 0
            || m[lf].compareTo(m[stepson])>= 0)
          break;
      }

      m[head] = m[stepson];

      head = stepson;
      int trail = Integer.numberOfTrailingZeros(p & ~1);
      p>>>= trail;
      pshift += trail;
      isTrusty = false;
    }

    if (!isTrusty) {
      m[head] = val;
      sift(m, pshift, head);
    }
  }