602 lines
24 KiB
Java
602 lines
24 KiB
Java
/*
|
|
* Copyright (C) 2011 The Android Open Source Project
|
|
*
|
|
* Licensed under the Apache License, Version 2.0 (the "License"); you may not
|
|
* use this file except in compliance with the License. You may obtain a copy of
|
|
* the License at
|
|
*
|
|
* http://www.apache.org/licenses/LICENSE-2.0
|
|
*
|
|
* Unless required by applicable law or agreed to in writing, software
|
|
* distributed under the License is distributed on an "AS IS" BASIS, WITHOUT
|
|
* WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the
|
|
* License for the specific language governing permissions and limitations under
|
|
* the License.
|
|
*/
|
|
|
|
package com.android.inputmethod.latin;
|
|
|
|
import java.util.ArrayList;
|
|
import java.util.Arrays;
|
|
import java.util.Collections;
|
|
import java.util.Iterator;
|
|
import java.util.LinkedList;
|
|
import java.util.List;
|
|
|
|
/**
|
|
* A dictionary that can fusion heads and tails of words for more compression.
|
|
*/
|
|
public class FusionDictionary implements Iterable<Word> {
|
|
|
|
/**
|
|
* A node of the dictionary, containing several CharGroups.
|
|
*
|
|
* A node is but an ordered array of CharGroups, which essentially contain all the
|
|
* real information.
|
|
* This class also contains fields to cache size and address, to help with binary
|
|
* generation.
|
|
*/
|
|
public static class Node {
|
|
ArrayList<CharGroup> mData;
|
|
// To help with binary generation
|
|
int mCachedSize;
|
|
int mCachedAddress;
|
|
public Node() {
|
|
mData = new ArrayList<CharGroup>();
|
|
mCachedSize = Integer.MIN_VALUE;
|
|
mCachedAddress = Integer.MIN_VALUE;
|
|
}
|
|
public Node(ArrayList<CharGroup> data) {
|
|
mData = data;
|
|
mCachedSize = Integer.MIN_VALUE;
|
|
mCachedAddress = Integer.MIN_VALUE;
|
|
}
|
|
}
|
|
|
|
/**
|
|
* A string with a frequency.
|
|
*
|
|
* This represents an "attribute", that is either a bigram or a shortcut.
|
|
*/
|
|
public static class WeightedString {
|
|
final String mWord;
|
|
final int mFrequency;
|
|
public WeightedString(String word, int frequency) {
|
|
mWord = word;
|
|
mFrequency = frequency;
|
|
}
|
|
}
|
|
|
|
/**
|
|
* A group of characters, with a frequency, shortcuts, bigrams, and children.
|
|
*
|
|
* This is the central class of the in-memory representation. A CharGroup is what can
|
|
* be seen as a traditional "trie node", except it can hold several characters at the
|
|
* same time. A CharGroup essentially represents one or several characters in the middle
|
|
* of the trie trie; as such, it can be a terminal, and it can have children.
|
|
* In this in-memory representation, whether the CharGroup is a terminal or not is represented
|
|
* in the frequency, where NOT_A_TERMINAL (= -1) means this is not a terminal and any other
|
|
* value is the frequency of this terminal. A terminal may have non-null shortcuts and/or
|
|
* bigrams, but a non-terminal may not. Moreover, children, if present, are null.
|
|
*/
|
|
public static class CharGroup {
|
|
public static final int NOT_A_TERMINAL = -1;
|
|
final int mChars[];
|
|
final ArrayList<WeightedString> mBigrams;
|
|
final int mFrequency; // NOT_A_TERMINAL == mFrequency indicates this is not a terminal.
|
|
Node mChildren;
|
|
// The two following members to help with binary generation
|
|
int mCachedSize;
|
|
int mCachedAddress;
|
|
|
|
public CharGroup(final int[] chars,
|
|
final ArrayList<WeightedString> bigrams, final int frequency) {
|
|
mChars = chars;
|
|
mFrequency = frequency;
|
|
mBigrams = bigrams;
|
|
mChildren = null;
|
|
}
|
|
|
|
public CharGroup(final int[] chars,
|
|
final ArrayList<WeightedString> bigrams, final int frequency, final Node children) {
|
|
mChars = chars;
|
|
mFrequency = frequency;
|
|
mBigrams = bigrams;
|
|
mChildren = children;
|
|
}
|
|
|
|
public void addChild(CharGroup n) {
|
|
if (null == mChildren) {
|
|
mChildren = new Node();
|
|
}
|
|
mChildren.mData.add(n);
|
|
}
|
|
|
|
public boolean isTerminal() {
|
|
return NOT_A_TERMINAL != mFrequency;
|
|
}
|
|
|
|
public boolean hasSeveralChars() {
|
|
assert(mChars.length > 0);
|
|
return 1 < mChars.length;
|
|
}
|
|
}
|
|
|
|
/**
|
|
* Options global to the dictionary.
|
|
*
|
|
* There are no options at the moment, so this class is empty.
|
|
*/
|
|
public static class DictionaryOptions {
|
|
}
|
|
|
|
|
|
public final DictionaryOptions mOptions;
|
|
public final Node mRoot;
|
|
|
|
public FusionDictionary() {
|
|
mOptions = new DictionaryOptions();
|
|
mRoot = new Node();
|
|
}
|
|
|
|
public FusionDictionary(final Node root, final DictionaryOptions options) {
|
|
mRoot = root;
|
|
mOptions = options;
|
|
}
|
|
|
|
/**
|
|
* Helper method to convert a String to an int array.
|
|
*/
|
|
static private int[] getCodePoints(String word) {
|
|
final int wordLength = word.length();
|
|
int[] array = new int[word.codePointCount(0, wordLength)];
|
|
for (int i = 0; i < wordLength; ++i) {
|
|
array[i] = word.codePointAt(i);
|
|
}
|
|
return array;
|
|
}
|
|
|
|
/**
|
|
* Helper method to add a word as a string.
|
|
*
|
|
* This method adds a word to the dictionary with the given frequency. Optional
|
|
* lists of bigrams and shortcuts can be passed here. For each word inside,
|
|
* they will be added to the dictionary as necessary.
|
|
*
|
|
* @param word the word to add.
|
|
* @param frequency the frequency of the word, in the range [0..255].
|
|
* @param bigrams a list of bigrams, or null.
|
|
*/
|
|
public void add(String word, int frequency, ArrayList<WeightedString> bigrams) {
|
|
if (null != bigrams) {
|
|
for (WeightedString bigram : bigrams) {
|
|
final CharGroup t = findWordInTree(mRoot, bigram.mWord);
|
|
if (null == t) {
|
|
add(getCodePoints(bigram.mWord), 0, null);
|
|
}
|
|
}
|
|
}
|
|
add(getCodePoints(word), frequency, bigrams);
|
|
}
|
|
|
|
/**
|
|
* Sanity check for a node.
|
|
*
|
|
* This method checks that all CharGroups in a node are ordered as expected.
|
|
* If they are, nothing happens. If they aren't, an exception is thrown.
|
|
*/
|
|
private void checkStack(Node node) {
|
|
ArrayList<CharGroup> stack = node.mData;
|
|
int lastValue = -1;
|
|
for (int i = 0; i < stack.size(); ++i) {
|
|
int currentValue = stack.get(i).mChars[0];
|
|
if (currentValue <= lastValue)
|
|
throw new RuntimeException("Invalid stack");
|
|
else
|
|
lastValue = currentValue;
|
|
}
|
|
}
|
|
|
|
/**
|
|
* Add a word to this dictionary.
|
|
*
|
|
* The bigrams, if any, have to be in the dictionary already. If they aren't,
|
|
* an exception is thrown.
|
|
*
|
|
* @param word the word, as an int array.
|
|
* @param frequency the frequency of the word, in the range [0..255].
|
|
* @param bigrams an optional list of bigrams for this word (null if none).
|
|
*/
|
|
private void add(int[] word, int frequency, ArrayList<WeightedString> bigrams) {
|
|
assert(frequency >= 0 && frequency <= 255);
|
|
Node currentNode = mRoot;
|
|
int charIndex = 0;
|
|
|
|
CharGroup currentGroup = null;
|
|
int differentCharIndex = 0; // Set by the loop to the index of the char that differs
|
|
int nodeIndex = findIndexOfChar(mRoot, word[charIndex]);
|
|
while (CHARACTER_NOT_FOUND != nodeIndex) {
|
|
currentGroup = currentNode.mData.get(nodeIndex);
|
|
differentCharIndex = compareArrays(currentGroup.mChars, word, charIndex);
|
|
if (ARRAYS_ARE_EQUAL != differentCharIndex
|
|
&& differentCharIndex < currentGroup.mChars.length) break;
|
|
if (null == currentGroup.mChildren) break;
|
|
charIndex += currentGroup.mChars.length;
|
|
if (charIndex >= word.length) break;
|
|
currentNode = currentGroup.mChildren;
|
|
nodeIndex = findIndexOfChar(currentNode, word[charIndex]);
|
|
}
|
|
|
|
if (-1 == nodeIndex) {
|
|
// No node at this point to accept the word. Create one.
|
|
final int insertionIndex = findInsertionIndex(currentNode, word[charIndex]);
|
|
final CharGroup newGroup = new CharGroup(
|
|
Arrays.copyOfRange(word, charIndex, word.length), bigrams, frequency);
|
|
currentNode.mData.add(insertionIndex, newGroup);
|
|
checkStack(currentNode);
|
|
} else {
|
|
// There is a word with a common prefix.
|
|
if (differentCharIndex == currentGroup.mChars.length) {
|
|
if (charIndex + differentCharIndex >= word.length) {
|
|
// The new word is a prefix of an existing word, but the node on which it
|
|
// should end already exists as is.
|
|
if (currentGroup.mFrequency > 0) {
|
|
throw new RuntimeException("Such a word already exists in the dictionary : "
|
|
+ new String(word, 0, word.length));
|
|
} else {
|
|
final CharGroup newNode = new CharGroup(currentGroup.mChars,
|
|
bigrams, frequency, currentGroup.mChildren);
|
|
currentNode.mData.set(nodeIndex, newNode);
|
|
checkStack(currentNode);
|
|
}
|
|
} else {
|
|
// The new word matches the full old word and extends past it.
|
|
// We only have to create a new node and add it to the end of this.
|
|
final CharGroup newNode = new CharGroup(
|
|
Arrays.copyOfRange(word, charIndex + differentCharIndex, word.length),
|
|
bigrams, frequency);
|
|
currentGroup.mChildren = new Node();
|
|
currentGroup.mChildren.mData.add(newNode);
|
|
}
|
|
} else {
|
|
if (0 == differentCharIndex) {
|
|
// Exact same word. Check the frequency is 0 or -1, and update.
|
|
if (0 != frequency) {
|
|
if (0 < currentGroup.mFrequency) {
|
|
throw new RuntimeException("This word already exists with frequency "
|
|
+ currentGroup.mFrequency + " : "
|
|
+ new String(word, 0, word.length));
|
|
}
|
|
final CharGroup newGroup = new CharGroup(word,
|
|
currentGroup.mBigrams, frequency, currentGroup.mChildren);
|
|
currentNode.mData.set(nodeIndex, newGroup);
|
|
}
|
|
} else {
|
|
// Partial prefix match only. We have to replace the current node with a node
|
|
// containing the current prefix and create two new ones for the tails.
|
|
Node newChildren = new Node();
|
|
final CharGroup newOldWord = new CharGroup(
|
|
Arrays.copyOfRange(currentGroup.mChars, differentCharIndex,
|
|
currentGroup.mChars.length),
|
|
currentGroup.mBigrams, currentGroup.mFrequency, currentGroup.mChildren);
|
|
newChildren.mData.add(newOldWord);
|
|
|
|
final CharGroup newParent;
|
|
if (charIndex + differentCharIndex >= word.length) {
|
|
newParent = new CharGroup(
|
|
Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex),
|
|
bigrams, frequency, newChildren);
|
|
} else {
|
|
newParent = new CharGroup(
|
|
Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex),
|
|
null, -1, newChildren);
|
|
final CharGroup newWord = new CharGroup(
|
|
Arrays.copyOfRange(word, charIndex + differentCharIndex,
|
|
word.length), bigrams, frequency);
|
|
final int addIndex = word[charIndex + differentCharIndex]
|
|
> currentGroup.mChars[differentCharIndex] ? 1 : 0;
|
|
newChildren.mData.add(addIndex, newWord);
|
|
}
|
|
currentNode.mData.set(nodeIndex, newParent);
|
|
}
|
|
checkStack(currentNode);
|
|
}
|
|
}
|
|
}
|
|
|
|
/**
|
|
* Custom comparison of two int arrays taken to contain character codes.
|
|
*
|
|
* This method compares the two arrays passed as an argument in a lexicographic way,
|
|
* with an offset in the dst string.
|
|
* This method does NOT test for the first character. It is taken to be equal.
|
|
* I repeat: this method starts the comparison at 1 <> dstOffset + 1.
|
|
* The index where the strings differ is returned. ARRAYS_ARE_EQUAL = 0 is returned if the
|
|
* strings are equal. This works BECAUSE we don't look at the first character.
|
|
*
|
|
* @param src the left-hand side string of the comparison.
|
|
* @param dst the right-hand side string of the comparison.
|
|
* @param dstOffset the offset in the right-hand side string.
|
|
* @return the index at which the strings differ, or ARRAYS_ARE_EQUAL = 0 if they don't.
|
|
*/
|
|
private static int ARRAYS_ARE_EQUAL = 0;
|
|
private static int compareArrays(final int[] src, final int[] dst, int dstOffset) {
|
|
// We do NOT test the first char, because we come from a method that already
|
|
// tested it.
|
|
for (int i = 1; i < src.length; ++i) {
|
|
if (dstOffset + i >= dst.length) return i;
|
|
if (src[i] != dst[dstOffset + i]) return i;
|
|
}
|
|
if (dst.length > src.length) return src.length;
|
|
return ARRAYS_ARE_EQUAL;
|
|
}
|
|
|
|
/**
|
|
* Helper class that compares and sorts two chargroups according to their
|
|
* first element only. I repeat: ONLY the first element is considered, the rest
|
|
* is ignored.
|
|
* This comparator imposes orderings that are inconsistent with equals.
|
|
*/
|
|
static private class CharGroupComparator implements java.util.Comparator {
|
|
public int compare(Object o1, Object o2) {
|
|
final CharGroup c1 = (CharGroup)o1;
|
|
final CharGroup c2 = (CharGroup)o2;
|
|
if (c1.mChars[0] == c2.mChars[0]) return 0;
|
|
return c1.mChars[0] < c2.mChars[0] ? -1 : 1;
|
|
}
|
|
public boolean equals(Object o) {
|
|
return o instanceof CharGroupComparator;
|
|
}
|
|
}
|
|
final static private CharGroupComparator CHARGROUP_COMPARATOR = new CharGroupComparator();
|
|
|
|
/**
|
|
* Finds the insertion index of a character within a node.
|
|
*/
|
|
private static int findInsertionIndex(final Node node, int character) {
|
|
final List data = node.mData;
|
|
final CharGroup reference = new CharGroup(new int[] { character }, null, 0);
|
|
int result = Collections.binarySearch(data, reference, CHARGROUP_COMPARATOR);
|
|
return result >= 0 ? result : -result - 1;
|
|
}
|
|
|
|
/**
|
|
* Find the index of a char in a node, if it exists.
|
|
*
|
|
* @param node the node to search in.
|
|
* @param character the character to search for.
|
|
* @return the position of the character if it's there, or CHARACTER_NOT_FOUND = -1 else.
|
|
*/
|
|
private static int CHARACTER_NOT_FOUND = -1;
|
|
private static int findIndexOfChar(final Node node, int character) {
|
|
final int insertionIndex = findInsertionIndex(node, character);
|
|
if (node.mData.size() <= insertionIndex) return CHARACTER_NOT_FOUND;
|
|
return character == node.mData.get(insertionIndex).mChars[0] ? insertionIndex
|
|
: CHARACTER_NOT_FOUND;
|
|
}
|
|
|
|
/**
|
|
* Helper method to find a word in a given branch.
|
|
*/
|
|
public static CharGroup findWordInTree(Node node, final String s) {
|
|
int index = 0;
|
|
final StringBuilder checker = new StringBuilder();
|
|
|
|
CharGroup currentGroup;
|
|
do {
|
|
int indexOfGroup = findIndexOfChar(node, s.codePointAt(index));
|
|
if (CHARACTER_NOT_FOUND == indexOfGroup) return null;
|
|
currentGroup = node.mData.get(indexOfGroup);
|
|
checker.append(new String(currentGroup.mChars, 0, currentGroup.mChars.length));
|
|
index += currentGroup.mChars.length;
|
|
if (index < s.length()) {
|
|
node = currentGroup.mChildren;
|
|
}
|
|
} while (null != node && index < s.length());
|
|
|
|
if (!s.equals(checker.toString())) return null;
|
|
return currentGroup;
|
|
}
|
|
|
|
/**
|
|
* Recursively count the number of character groups in a given branch of the trie.
|
|
*
|
|
* @param node the parent node.
|
|
* @return the number of char groups in all the branch under this node.
|
|
*/
|
|
public static int countCharGroups(final Node node) {
|
|
final int nodeSize = node.mData.size();
|
|
int size = nodeSize;
|
|
for (int i = nodeSize - 1; i >= 0; --i) {
|
|
CharGroup group = node.mData.get(i);
|
|
if (null != group.mChildren)
|
|
size += countCharGroups(group.mChildren);
|
|
}
|
|
return size;
|
|
}
|
|
|
|
/**
|
|
* Recursively count the number of nodes in a given branch of the trie.
|
|
*
|
|
* @param node the node to count.
|
|
* @result the number of nodes in this branch.
|
|
*/
|
|
public static int countNodes(final Node node) {
|
|
int size = 1;
|
|
for (int i = node.mData.size() - 1; i >= 0; --i) {
|
|
CharGroup group = node.mData.get(i);
|
|
if (null != group.mChildren)
|
|
size += countNodes(group.mChildren);
|
|
}
|
|
return size;
|
|
}
|
|
|
|
// Historically, the tails of the words were going to be merged to save space.
|
|
// However, that would prevent the code to search for a specific address in log(n)
|
|
// time so this was abandoned.
|
|
// The code is still of interest as it does add some compression to any dictionary
|
|
// that has no need for attributes. Implementations that does not read attributes should be
|
|
// able to read a dictionary with merged tails.
|
|
// Also, the following code does support frequencies, as in, it will only merges
|
|
// tails that share the same frequency. Though it would result in the above loss of
|
|
// performance while searching by address, it is still technically possible to merge
|
|
// tails that contain attributes, but this code does not take that into account - it does
|
|
// not compare attributes and will merge terminals with different attributes regardless.
|
|
public void mergeTails() {
|
|
MakedictLog.i("Do not merge tails");
|
|
return;
|
|
|
|
// MakedictLog.i("Merging nodes. Number of nodes : " + countNodes(root));
|
|
// MakedictLog.i("Number of groups : " + countCharGroups(root));
|
|
//
|
|
// final HashMap<String, ArrayList<Node>> repository =
|
|
// new HashMap<String, ArrayList<Node>>();
|
|
// mergeTailsInner(repository, root);
|
|
//
|
|
// MakedictLog.i("Number of different pseudohashes : " + repository.size());
|
|
// int size = 0;
|
|
// for (ArrayList<Node> a : repository.values()) {
|
|
// size += a.size();
|
|
// }
|
|
// MakedictLog.i("Number of nodes after merge : " + (1 + size));
|
|
// MakedictLog.i("Recursively seen nodes : " + countNodes(root));
|
|
}
|
|
|
|
// The following methods are used by the deactivated mergeTails()
|
|
// private static boolean isEqual(Node a, Node b) {
|
|
// if (null == a && null == b) return true;
|
|
// if (null == a || null == b) return false;
|
|
// if (a.data.size() != b.data.size()) return false;
|
|
// final int size = a.data.size();
|
|
// for (int i = size - 1; i >= 0; --i) {
|
|
// CharGroup aGroup = a.data.get(i);
|
|
// CharGroup bGroup = b.data.get(i);
|
|
// if (aGroup.frequency != bGroup.frequency) return false;
|
|
// if (aGroup.alternates == null && bGroup.alternates != null) return false;
|
|
// if (aGroup.alternates != null && !aGroup.equals(bGroup.alternates)) return false;
|
|
// if (!Arrays.equals(aGroup.chars, bGroup.chars)) return false;
|
|
// if (!isEqual(aGroup.children, bGroup.children)) return false;
|
|
// }
|
|
// return true;
|
|
// }
|
|
|
|
// static private HashMap<String, ArrayList<Node>> mergeTailsInner(
|
|
// final HashMap<String, ArrayList<Node>> map, final Node node) {
|
|
// final ArrayList<CharGroup> branches = node.data;
|
|
// final int nodeSize = branches.size();
|
|
// for (int i = 0; i < nodeSize; ++i) {
|
|
// CharGroup group = branches.get(i);
|
|
// if (null != group.children) {
|
|
// String pseudoHash = getPseudoHash(group.children);
|
|
// ArrayList<Node> similarList = map.get(pseudoHash);
|
|
// if (null == similarList) {
|
|
// similarList = new ArrayList<Node>();
|
|
// map.put(pseudoHash, similarList);
|
|
// }
|
|
// boolean merged = false;
|
|
// for (Node similar : similarList) {
|
|
// if (isEqual(group.children, similar)) {
|
|
// group.children = similar;
|
|
// merged = true;
|
|
// break;
|
|
// }
|
|
// }
|
|
// if (!merged) {
|
|
// similarList.add(group.children);
|
|
// }
|
|
// mergeTailsInner(map, group.children);
|
|
// }
|
|
// }
|
|
// return map;
|
|
// }
|
|
|
|
// private static String getPseudoHash(final Node node) {
|
|
// StringBuilder s = new StringBuilder();
|
|
// for (CharGroup g : node.data) {
|
|
// s.append(g.frequency);
|
|
// for (int ch : g.chars){
|
|
// s.append(Character.toChars(ch));
|
|
// }
|
|
// }
|
|
// return s.toString();
|
|
// }
|
|
|
|
/**
|
|
* Iterator to walk through a dictionary.
|
|
*
|
|
* This is purely for convenience.
|
|
*/
|
|
public static class DictionaryIterator implements Iterator<Word> {
|
|
|
|
private static class Position {
|
|
public Iterator<CharGroup> pos;
|
|
public int length;
|
|
public Position(ArrayList<CharGroup> groups) {
|
|
pos = groups.iterator();
|
|
length = 0;
|
|
}
|
|
}
|
|
final StringBuilder mCurrentString;
|
|
final LinkedList<Position> mPositions;
|
|
|
|
public DictionaryIterator(ArrayList<CharGroup> root) {
|
|
mCurrentString = new StringBuilder();
|
|
mPositions = new LinkedList<Position>();
|
|
final Position rootPos = new Position(root);
|
|
mPositions.add(rootPos);
|
|
}
|
|
|
|
@Override
|
|
public boolean hasNext() {
|
|
for (Position p : mPositions) {
|
|
if (p.pos.hasNext()) {
|
|
return true;
|
|
}
|
|
}
|
|
return false;
|
|
}
|
|
|
|
@Override
|
|
public Word next() {
|
|
Position currentPos = mPositions.getLast();
|
|
mCurrentString.setLength(mCurrentString.length() - currentPos.length);
|
|
|
|
do {
|
|
if (currentPos.pos.hasNext()) {
|
|
final CharGroup currentGroup = currentPos.pos.next();
|
|
currentPos.length = currentGroup.mChars.length;
|
|
for (int i : currentGroup.mChars)
|
|
mCurrentString.append(Character.toChars(i));
|
|
if (null != currentGroup.mChildren) {
|
|
currentPos = new Position(currentGroup.mChildren.mData);
|
|
mPositions.addLast(currentPos);
|
|
}
|
|
if (currentGroup.mFrequency >= 0)
|
|
return new Word(mCurrentString.toString(), currentGroup.mFrequency,
|
|
currentGroup.mBigrams);
|
|
} else {
|
|
mPositions.removeLast();
|
|
currentPos = mPositions.getLast();
|
|
mCurrentString.setLength(mCurrentString.length() - mPositions.getLast().length);
|
|
}
|
|
} while(true);
|
|
}
|
|
|
|
@Override
|
|
public void remove() {
|
|
throw new UnsupportedOperationException("Unsupported yet");
|
|
}
|
|
|
|
}
|
|
|
|
/**
|
|
* Method to return an iterator.
|
|
*
|
|
* This method enables Java's enhanced for loop. With this you can have a FusionDictionary x
|
|
* and say : for (Word w : x) {}
|
|
*/
|
|
@Override
|
|
public Iterator<Word> iterator() {
|
|
return new DictionaryIterator(mRoot.mData);
|
|
}
|
|
}
|