http://www.keithschwarz.com/interesting/code/?dir=vlist
* An implementation of the List abstraction backed by a VList. The VList is
* similar to a standard dynamic array implementation, except that the old
* arrays are chained together to form the implementation, rather than
* discarded. For example, here is one possible VList holding six elements:
*
* <pre>
*
* +------+ +------+ +------+
* | next | --} | next | --} | next | -| nil
* +------+ +------+ +------+
* nil |- | prev | {-- | prev | {-- | prev |
* +------+ +------+ +------+
* | | | 2 | | 0 |
* +------+ +------+ +------+
* | 5 | | 1 |
* +------+ +------+
* | 4 |
* +------+
* | 3 |
* +------+
*
* </pre>
*
* Each column represents a linked list cell containing a fixed-size array. The
* size of this array depends on which cell is being referenced. In particular,
* the nth cell from the end has 2^n elements in it. In this manner, for a
* uniformly-randomly chosen position, the expected time to look up that element
* is O(1), though in the worst case it is O(log n). This random access is
* significantly better than a linked list, though in the worst case is not as
* good as a dynamic array.
* <p>
* Notice that in each array, elements grow up, with the lower-numbered indices
* of the array holding elements in the array whose positions are higher. This
* makes it easier to look up individual elements.
* <p>
*
* @author Keith Schwarz (htiek@cs.stanford.edu)
*/
public final class VList<T> extends AbstractList<T> {
/**
* A single cell in the VList implementation.
*/
private static final class VListCell<T> {
public final T[] mElems;
public final VListCell<T> mNext;
/* This field is not mutable because when new elements are added/deleted
* from the main list, the previous pointer needs to be updated.
* However, next links never change because the list only grows in one
* direction.
*/
public VListCell<T> mPrev;
/* The number of unused elements in this cell. Alternatively, you can
* think of this as the index in the array in which the first used
* element appears. Both interpretations are used in this
* implementation.
*/
public int mFreeSpace;
/**
* Constructs a new VListCell with the specified number of elements and
* specified next element.
*
* @param numElems
* The number of elements this cell should have space for.
* @param next
* The cell in the list of cells that follows this one.
*/
public VListCell(int numElems, VListCell<T> next) {
mElems = (T[]) new Object[numElems];
mNext = next;
mPrev = null;
/* Update the next cell to point back to us. */
if (next != null)
next.mPrev = this;
/* We have free space equal to the number of elements. */
mFreeSpace = numElems;
}
}
/**
* A utility struct containing information about where an element is in the
* VList. Methods that need to manipulate individual elements of the list
* use this struct to communicate where in the list to look for that
* element.
*/
private static final class VListLocation<T> {
public final VListCell<T> mCell;
public final int mOffset;
public VListLocation(VListCell<T> cell, int offset) {
mCell = cell;
mOffset = offset;
}
}
/* Pointer to the head of the VList, which contains the final elements of
* the list.
*/
private VListCell<T> mHead;
/* Cached total number of elements in the array. */
private int mSize;
/**
* Adds a new element to the end of the array.
*
* @param elem
* The element to add.
* @return true
*/
@Override
public boolean add(T elem) {
/* If no free space exists, add a new element to the list. */
if (mHead == null || mHead.mFreeSpace == 0)
mHead = new VListCell<T>(mHead == null? 1 : mHead.mElems.length * 2, mHead);
/* Prepend this element to the current cell. */
mHead.mElems[(mHead.mFreeSpace--) - 1] = elem;
++mSize;
/* Success! */
return true;
}
/**
* Given an absolute offset into the VList, returns an object describing
* where that object is in the VList.
*
* @param index
* The index into the VList.
* @return A VListLocation object holding information about where that
* element can be found.
*/
private VListLocation<T> locateElement(int index) {
/* Bounds-check. */
if (index >= size() || index < 0)
throw new IndexOutOfBoundsException("Position " + index + "; size "
+ size());
/* Because the list is stored with new elements in front and old
* elements in back, we'll invert the index so that 0 refers to the
* final element of the array and size() - 1 refers to the first
* element.
*/
index = size() - 1 - index;
/* Scan across the cells, looking for the first one that can hold our
* entry. We do this by continuously skipping cells until we find one
* that can be sure to hold this element.
*
* Note that each cell has mElems.length elements, of which mFreeSpace
* is used. This means that the total number of used elements in each
* cell is mElems.length - mFreeSpace.
*/
VListCell<T> curr = mHead;
while (index >= curr.mElems.length - curr.mFreeSpace) {
/* Skip past all these elements. */
index -= curr.mElems.length - curr.mFreeSpace;
curr = curr.mNext;
}
/* We're now in the correct location for what we need to do. The element
* we want can be found by indexing the proper amount beyond the free
* space.
*/
return new VListLocation<T>(curr, index + curr.mFreeSpace);
}
/**
* Scans for the proper location in the cell list for the element, then
* returns the element at that position.
*
* @param index
* The index at which to look up the element.
* @return The element at that position.
*/
@Override
public T get(int index) {
VListLocation<T> where = locateElement(index);
/* Return the element in the current position of this array. */
return where.mCell.mElems[where.mOffset];
}
/**
* Returns the cached size.
*
* @return The size of the VList.
*/
@Override
public int size() {
return mSize;
}
/**
* Sets an element at a particular position to have a particular value.
*
* @param index
* The index at which to write a new value.
* @param value
* The value to write at that position.
* @return The value originally held at that position.
*/
@Override
public T set(int index, T value) {
VListLocation<T> where = locateElement(index);
/* Cache the element in the current position of this array. */
T result = where.mCell.mElems[where.mOffset];
where.mCell.mElems[where.mOffset] = value;
return result;
}
/**
* Removes the element at the specified position from the VList, returning
* its value.
*
* @param index
* The index at which the element should be removed.
* @return The value held at that position.
*/
@Override
public T remove(int index) {
VListLocation<T> where = locateElement(index);
/* Cache the value that will be removed. */
T result = where.mCell.mElems[where.mOffset];
/* Invoke the helper to do most of the work. */
removeAtPosition(where);
return result;
}
/**
* Removes the element at the indicated VListLocation.
*
* @param where
* The location at which the element should be removed.
*/
private void removeAtPosition(VListLocation<T> where) {
/* Scan backward across the blocks after this element, shuffling array
* elements down a position and copying the last element of the next
* block over to fill in the top.
*
* The variable shuffleTargetPosition indicates the first element of the
* block that should be overwritten during the shuffle-down. In the
* first block, this is the position of the element that was
* overwritten. In all other blocks, it's the last element.
*/
VListCell<T> curr = where.mCell;
for (int shuffleTargetPosition = where.mOffset; curr != null;
curr = curr.mPrev, shuffleTargetPosition = (curr == null? 0 : curr.mElems.length - 1)) {
/* Shuffle down each element in the current array on top of the
* target position. Note that in the final block, this may end up
* copying a whole bunch of null values down. This is more work than
* necessary, but is harmless and doesn't change the asymptotic
* runtime (since the last block has size O(n)).
*/
for (int i = shuffleTargetPosition - 1; i >= 0; --i)
curr.mElems[i + 1] = curr.mElems[i];
/* Copy the last element of the next array to the top of this array,
* unless this is the first block (in which case there is no next
* array).
*/
if (curr.mPrev != null)
curr.mElems[0] = curr.mPrev.mElems[curr.mPrev.mElems.length - 1];
}
/* The head just lost an element, so it has some more free space. Null
* out the lost element and increase the free space.
*/
++mHead.mFreeSpace;
mHead.mElems[mHead.mFreeSpace - 1] = null;
/* The whole list just lost an element. */
--mSize;
/* If the head is entirely free, remove it from the list. */
if (mHead.mFreeSpace == mHead.mElems.length) {
mHead = mHead.mNext;
/* If there is at least one block left, remove the previous block
* from the linked list.
*/
if (mHead != null)
mHead.mPrev = null;
}
}
/**
* A custom iterator class that traverses the elements of this container in
* an intelligent way. The normal iterator will call get repeatedly, which
* is slow because it has to continuously scan for the proper location of
* the next element. This iterator works by traversing the cells as a proper
* linked list.
*/
private final class VListIterator implements Iterator<T> {
/* The cell and position in that cell that we are about to visit. We
* maintain the invariant that if there is a next element, mCurrCell is
* non-null and conversely that if mCurrCell is null, there is no next
* element.
*/
private VListCell<T> mCurrCell;
private int mCurrIndex;
/* Stores whether we have something to remove (i.e. whether we've called
* next() without an invervening remove()).
*/
private boolean mCanRemove;
/**
* Constructs a new VListIterator that will traverse the elements of the
* containing VList.
*/
public VListIterator() {
/* Scan to the tail using the "pointer chase" algorithm. When this
* terminates, prev will hold a pointer to the last element of the
* list.
*/
VListCell<T> curr, prev;
for (curr = mHead, prev = null; curr != null; prev = curr, curr = curr.mNext)
;
/* Set the current cell to the tail. */
mCurrCell = prev;
/* If the tail isn't null, it must be a full list of size 1. Set the
* current index appropriately.
*/
if (mCurrCell != null)
mCurrIndex = 0;
}
/**
* As per our invariant, returns whether mCurrCell is non-null.
*/
@Override
public boolean hasNext() {
return mCurrCell != null;
}
/**
* Advances the iterator and returns the element it used to be over.
*/
@Override
public T next() {
/* Bounds-check. */
if (!hasNext())
throw new NoSuchElementException();
/* Cache the return value; we'll be moving off of it soon. */
T result = mCurrCell.mElems[mCurrIndex];
/* Back up one step. */
--mCurrIndex;
/* If we walked off the end of the buffer, advance to the next
* element of the list.
*/
if (mCurrIndex < mCurrCell.mFreeSpace) {
mCurrCell = mCurrCell.mPrev;
/* Update the next get location, provided of course that we
* didn't just walk off the end of the list.
*/
if (mCurrCell != null)
mCurrIndex = mCurrCell.mElems.length - 1;
}
/* Since there was indeed an element, we can remove it. */
mCanRemove = true;
return result;
}
/**
* Removes the last element we visited.
*/
@Override
public void remove() {
/* Check whether there's something to remove. */
if (!mCanRemove)
throw new IllegalStateException(
"remove() without next(), or double remove().");
/* Clear the flag saying we can do this. */
mCanRemove = false;
/* There are several cases to consider. If the current cell is null,
* we've walked off the end of the array, so we want to remove the
* very last element. If the current cell isn't null and the cursor
* is in the middle, remove the previous element and back up a step.
* If the current cell isn't null and the cursor is at the front,
* remove the element one step before us and back up a step.
*/
/* Case 1. */
if (mCurrCell == null)
VList.this.remove(size() - 1);
/* Case 2. */
else if (mCurrIndex != mCurrCell.mElems.length - 1) {
/* Back up a step, and remove the element at this position.
* After the remove completes, the element here should be the
* next element to visit.
*/
++mCurrIndex;
removeAtPosition(new VListLocation<T>(mCurrCell, mCurrIndex));
}
/* Case 3. */
else {
/* Back up a step to the top of the previous list. We know that
* the top will be at position 0, since all internal blocks are
* completely full. We also know that we aren't at the very
* front of the list, since if we were, then the call to next()
* that enabled this call would have pushed us to the next
* location.
*/
mCurrCell = mCurrCell.mNext;
mCurrIndex = 0;
removeAtPosition(new VListLocation<T>(mCurrCell, mCurrIndex));
}
}
}
/**
* Returns a custom iterator rather than the default.
*/
@Override
public Iterator<T> iterator() {
return new VListIterator();
}
}
http://www.sanfoundry.com/java-program-implement-vlist/
https://en.wikipedia.org/wiki/VList
the VList is a persistent data structure designed by Phil Bagwell in 2002 that combines the fast indexing of arrays with the easy extension of cons-based (or singly linked) linked lists.
https://en.wikipedia.org/wiki/Persistent_data_structure
a persistent data structure is a data structure that always preserves the previous version of itself when it is modified. Such data structures are effectively immutable, as their operations do not (visibly) update the structure in-place, but instead always yield a new updated structure.
* An implementation of the List abstraction backed by a VList. The VList is
* similar to a standard dynamic array implementation, except that the old
* arrays are chained together to form the implementation, rather than
* discarded. For example, here is one possible VList holding six elements:
*
* <pre>
*
* +------+ +------+ +------+
* | next | --} | next | --} | next | -| nil
* +------+ +------+ +------+
* nil |- | prev | {-- | prev | {-- | prev |
* +------+ +------+ +------+
* | | | 2 | | 0 |
* +------+ +------+ +------+
* | 5 | | 1 |
* +------+ +------+
* | 4 |
* +------+
* | 3 |
* +------+
*
* </pre>
*
* Each column represents a linked list cell containing a fixed-size array. The
* size of this array depends on which cell is being referenced. In particular,
* the nth cell from the end has 2^n elements in it. In this manner, for a
* uniformly-randomly chosen position, the expected time to look up that element
* is O(1), though in the worst case it is O(log n). This random access is
* significantly better than a linked list, though in the worst case is not as
* good as a dynamic array.
* <p>
* Notice that in each array, elements grow up, with the lower-numbered indices
* of the array holding elements in the array whose positions are higher. This
* makes it easier to look up individual elements.
* <p>
*
* @author Keith Schwarz (htiek@cs.stanford.edu)
*/
public final class VList<T> extends AbstractList<T> {
/**
* A single cell in the VList implementation.
*/
private static final class VListCell<T> {
public final T[] mElems;
public final VListCell<T> mNext;
/* This field is not mutable because when new elements are added/deleted
* from the main list, the previous pointer needs to be updated.
* However, next links never change because the list only grows in one
* direction.
*/
public VListCell<T> mPrev;
/* The number of unused elements in this cell. Alternatively, you can
* think of this as the index in the array in which the first used
* element appears. Both interpretations are used in this
* implementation.
*/
public int mFreeSpace;
/**
* Constructs a new VListCell with the specified number of elements and
* specified next element.
*
* @param numElems
* The number of elements this cell should have space for.
* @param next
* The cell in the list of cells that follows this one.
*/
public VListCell(int numElems, VListCell<T> next) {
mElems = (T[]) new Object[numElems];
mNext = next;
mPrev = null;
/* Update the next cell to point back to us. */
if (next != null)
next.mPrev = this;
/* We have free space equal to the number of elements. */
mFreeSpace = numElems;
}
}
/**
* A utility struct containing information about where an element is in the
* VList. Methods that need to manipulate individual elements of the list
* use this struct to communicate where in the list to look for that
* element.
*/
private static final class VListLocation<T> {
public final VListCell<T> mCell;
public final int mOffset;
public VListLocation(VListCell<T> cell, int offset) {
mCell = cell;
mOffset = offset;
}
}
/* Pointer to the head of the VList, which contains the final elements of
* the list.
*/
private VListCell<T> mHead;
/* Cached total number of elements in the array. */
private int mSize;
/**
* Adds a new element to the end of the array.
*
* @param elem
* The element to add.
* @return true
*/
@Override
public boolean add(T elem) {
/* If no free space exists, add a new element to the list. */
if (mHead == null || mHead.mFreeSpace == 0)
mHead = new VListCell<T>(mHead == null? 1 : mHead.mElems.length * 2, mHead);
/* Prepend this element to the current cell. */
mHead.mElems[(mHead.mFreeSpace--) - 1] = elem;
++mSize;
/* Success! */
return true;
}
/**
* Given an absolute offset into the VList, returns an object describing
* where that object is in the VList.
*
* @param index
* The index into the VList.
* @return A VListLocation object holding information about where that
* element can be found.
*/
private VListLocation<T> locateElement(int index) {
/* Bounds-check. */
if (index >= size() || index < 0)
throw new IndexOutOfBoundsException("Position " + index + "; size "
+ size());
/* Because the list is stored with new elements in front and old
* elements in back, we'll invert the index so that 0 refers to the
* final element of the array and size() - 1 refers to the first
* element.
*/
index = size() - 1 - index;
/* Scan across the cells, looking for the first one that can hold our
* entry. We do this by continuously skipping cells until we find one
* that can be sure to hold this element.
*
* Note that each cell has mElems.length elements, of which mFreeSpace
* is used. This means that the total number of used elements in each
* cell is mElems.length - mFreeSpace.
*/
VListCell<T> curr = mHead;
while (index >= curr.mElems.length - curr.mFreeSpace) {
/* Skip past all these elements. */
index -= curr.mElems.length - curr.mFreeSpace;
curr = curr.mNext;
}
/* We're now in the correct location for what we need to do. The element
* we want can be found by indexing the proper amount beyond the free
* space.
*/
return new VListLocation<T>(curr, index + curr.mFreeSpace);
}
/**
* Scans for the proper location in the cell list for the element, then
* returns the element at that position.
*
* @param index
* The index at which to look up the element.
* @return The element at that position.
*/
@Override
public T get(int index) {
VListLocation<T> where = locateElement(index);
/* Return the element in the current position of this array. */
return where.mCell.mElems[where.mOffset];
}
/**
* Returns the cached size.
*
* @return The size of the VList.
*/
@Override
public int size() {
return mSize;
}
/**
* Sets an element at a particular position to have a particular value.
*
* @param index
* The index at which to write a new value.
* @param value
* The value to write at that position.
* @return The value originally held at that position.
*/
@Override
public T set(int index, T value) {
VListLocation<T> where = locateElement(index);
/* Cache the element in the current position of this array. */
T result = where.mCell.mElems[where.mOffset];
where.mCell.mElems[where.mOffset] = value;
return result;
}
/**
* Removes the element at the specified position from the VList, returning
* its value.
*
* @param index
* The index at which the element should be removed.
* @return The value held at that position.
*/
@Override
public T remove(int index) {
VListLocation<T> where = locateElement(index);
/* Cache the value that will be removed. */
T result = where.mCell.mElems[where.mOffset];
/* Invoke the helper to do most of the work. */
removeAtPosition(where);
return result;
}
/**
* Removes the element at the indicated VListLocation.
*
* @param where
* The location at which the element should be removed.
*/
private void removeAtPosition(VListLocation<T> where) {
/* Scan backward across the blocks after this element, shuffling array
* elements down a position and copying the last element of the next
* block over to fill in the top.
*
* The variable shuffleTargetPosition indicates the first element of the
* block that should be overwritten during the shuffle-down. In the
* first block, this is the position of the element that was
* overwritten. In all other blocks, it's the last element.
*/
VListCell<T> curr = where.mCell;
for (int shuffleTargetPosition = where.mOffset; curr != null;
curr = curr.mPrev, shuffleTargetPosition = (curr == null? 0 : curr.mElems.length - 1)) {
/* Shuffle down each element in the current array on top of the
* target position. Note that in the final block, this may end up
* copying a whole bunch of null values down. This is more work than
* necessary, but is harmless and doesn't change the asymptotic
* runtime (since the last block has size O(n)).
*/
for (int i = shuffleTargetPosition - 1; i >= 0; --i)
curr.mElems[i + 1] = curr.mElems[i];
/* Copy the last element of the next array to the top of this array,
* unless this is the first block (in which case there is no next
* array).
*/
if (curr.mPrev != null)
curr.mElems[0] = curr.mPrev.mElems[curr.mPrev.mElems.length - 1];
}
/* The head just lost an element, so it has some more free space. Null
* out the lost element and increase the free space.
*/
++mHead.mFreeSpace;
mHead.mElems[mHead.mFreeSpace - 1] = null;
/* The whole list just lost an element. */
--mSize;
/* If the head is entirely free, remove it from the list. */
if (mHead.mFreeSpace == mHead.mElems.length) {
mHead = mHead.mNext;
/* If there is at least one block left, remove the previous block
* from the linked list.
*/
if (mHead != null)
mHead.mPrev = null;
}
}
/**
* A custom iterator class that traverses the elements of this container in
* an intelligent way. The normal iterator will call get repeatedly, which
* is slow because it has to continuously scan for the proper location of
* the next element. This iterator works by traversing the cells as a proper
* linked list.
*/
private final class VListIterator implements Iterator<T> {
/* The cell and position in that cell that we are about to visit. We
* maintain the invariant that if there is a next element, mCurrCell is
* non-null and conversely that if mCurrCell is null, there is no next
* element.
*/
private VListCell<T> mCurrCell;
private int mCurrIndex;
/* Stores whether we have something to remove (i.e. whether we've called
* next() without an invervening remove()).
*/
private boolean mCanRemove;
/**
* Constructs a new VListIterator that will traverse the elements of the
* containing VList.
*/
public VListIterator() {
/* Scan to the tail using the "pointer chase" algorithm. When this
* terminates, prev will hold a pointer to the last element of the
* list.
*/
VListCell<T> curr, prev;
for (curr = mHead, prev = null; curr != null; prev = curr, curr = curr.mNext)
;
/* Set the current cell to the tail. */
mCurrCell = prev;
/* If the tail isn't null, it must be a full list of size 1. Set the
* current index appropriately.
*/
if (mCurrCell != null)
mCurrIndex = 0;
}
/**
* As per our invariant, returns whether mCurrCell is non-null.
*/
@Override
public boolean hasNext() {
return mCurrCell != null;
}
/**
* Advances the iterator and returns the element it used to be over.
*/
@Override
public T next() {
/* Bounds-check. */
if (!hasNext())
throw new NoSuchElementException();
/* Cache the return value; we'll be moving off of it soon. */
T result = mCurrCell.mElems[mCurrIndex];
/* Back up one step. */
--mCurrIndex;
/* If we walked off the end of the buffer, advance to the next
* element of the list.
*/
if (mCurrIndex < mCurrCell.mFreeSpace) {
mCurrCell = mCurrCell.mPrev;
/* Update the next get location, provided of course that we
* didn't just walk off the end of the list.
*/
if (mCurrCell != null)
mCurrIndex = mCurrCell.mElems.length - 1;
}
/* Since there was indeed an element, we can remove it. */
mCanRemove = true;
return result;
}
/**
* Removes the last element we visited.
*/
@Override
public void remove() {
/* Check whether there's something to remove. */
if (!mCanRemove)
throw new IllegalStateException(
"remove() without next(), or double remove().");
/* Clear the flag saying we can do this. */
mCanRemove = false;
/* There are several cases to consider. If the current cell is null,
* we've walked off the end of the array, so we want to remove the
* very last element. If the current cell isn't null and the cursor
* is in the middle, remove the previous element and back up a step.
* If the current cell isn't null and the cursor is at the front,
* remove the element one step before us and back up a step.
*/
/* Case 1. */
if (mCurrCell == null)
VList.this.remove(size() - 1);
/* Case 2. */
else if (mCurrIndex != mCurrCell.mElems.length - 1) {
/* Back up a step, and remove the element at this position.
* After the remove completes, the element here should be the
* next element to visit.
*/
++mCurrIndex;
removeAtPosition(new VListLocation<T>(mCurrCell, mCurrIndex));
}
/* Case 3. */
else {
/* Back up a step to the top of the previous list. We know that
* the top will be at position 0, since all internal blocks are
* completely full. We also know that we aren't at the very
* front of the list, since if we were, then the call to next()
* that enabled this call would have pushed us to the next
* location.
*/
mCurrCell = mCurrCell.mNext;
mCurrIndex = 0;
removeAtPosition(new VListLocation<T>(mCurrCell, mCurrIndex));
}
}
}
/**
* Returns a custom iterator rather than the default.
*/
@Override
public Iterator<T> iterator() {
return new VListIterator();
}
}
http://www.sanfoundry.com/java-program-implement-vlist/
https://en.wikipedia.org/wiki/VList
the VList is a persistent data structure designed by Phil Bagwell in 2002 that combines the fast indexing of arrays with the easy extension of cons-based (or singly linked) linked lists.
https://en.wikipedia.org/wiki/Persistent_data_structure
a persistent data structure is a data structure that always preserves the previous version of itself when it is modified. Such data structures are effectively immutable, as their operations do not (visibly) update the structure in-place, but instead always yield a new updated structure.
