Пример #1
0
        private void Add(IList <elementType> key, resultType value, int keyPos)
        {
            int size = key.Count - keyPos;

            if (size > 0)
            {
                //add the node needed if nessesary
                if (!nodes.ContainsKey(key[0]))
                {
                    DSTNode <elementType, resultType> next = new DSTNode <elementType, resultType>();
                    next.init();
                    nodes.Add(key[0], next);
                }

                nodes[key[0]].Add(key, value, keyPos++);
            }
            else
            {
                if (result == null)
                {
                    result = new List <resultType>();
                    result.Add(value);
                    count++;
                }
                else
                {
                    throw new ArgumentException("An element with the same key already exists in the DST.");
                }
            }
        }
Пример #2
0
        /// <summary>
        /// Get the item at an index.
        /// NOTE: this should be the only way to map a index to an item.
        /// All operations that are index bound must call this. This insures that there is only one key to item mapping in place.
        /// </summary>
        /// <param name="index">The index.</param>
        /// <param name="count">The count.</param>
        /// <param name="key">The key.</param>
        /// <param name="value">The value.</param>
        /// <returns></returns>
        private bool ItemAtIndex(int index, int count, ref LinkedList <elementType> key, out resultType value)
        {
            int c = count;

            Dictionary <elementType, DSTNode <elementType, resultType> > .KeyCollection keyList = nodes.Keys;
            List <elementType> sortedKeyList = new List <elementType>(keyList);

            //keyList.CopyTo(sortedKeyList, 0);
            sortedKeyList.Sort();

            foreach (elementType keyItem in sortedKeyList)
            {
                DSTNode <elementType, resultType> n = nodes[keyItem];
                if ((c + n.count) >= index)
                {
                    if (count == index)
                    {
                        //found a node whose final child is the one.
                        if (result != null)
                        {
                            value = this.result[0];
                            return(true);
                        }
                    }


                    if (ItemAtIndex(index, c, ref key, out value))
                    {
                        key.AddFirst(keyItem);
                        return(true);
                    }
                }

                c += n.count;
            }

            value = default(resultType);
            return(false);
        }
Пример #3
0
 public DST()
 {
     rootNode = new DSTNode <elementType, resultType>();
     rootNode.init();
 }