com.ctc.wstx.util.WordResolver Maven / Gradle / Ivy
Show all versions of webservices-rt Show documentation
package com.ctc.wstx.util;
import java.util.*;
/**
* A specialized Map/Symbol table - like data structure that can be used
* for both checking whether a word (passed in as a char array) exists
* in certain set of words AND getting that word as a String.
* It is reasonably efficient both time and speed-wise, at least for
* certain use cases; specifically, if there is no existing key to use,
* it is more efficient way to get to a shared copy of that String
* The general usage pattern is expected
* to be such that most checks are positive, ie. that the word indeed
* is contained in the structure.
*
* Although this is an efficient data struct for specific set of usage
* patterns, one restriction is that the full set of words to include has to
* be known before constructing the instnace. Also, the size of the set is
* limited to total word content of about 20k characters.
*
* TODO: Should document the internal data structure...
*/
public final class WordResolver
{
/**
* Maximum number of words (Strings) an instance can contain
*/
public final static int MAX_WORDS = 0x2000;
final static char CHAR_NULL = (char) 0;
/**
* Offset added to numbers to mark 'negative' numbers. Asymmetric,
* since range of negative markers needed is smaller than positive
* numbers...
*/
final static int NEGATIVE_OFFSET = 0x10000 - MAX_WORDS;
/**
* This is actually just a guess; but in general linear search should
* be faster for short sequences (definitely for 4 or less; maybe up
* to 8 or less?)
*/
final static int MIN_BINARY_SEARCH = 7;
/**
* Compressed presentation of the word set.
*/
final char[] mData;
/**
* Array of actual words returned resolved for matches.
*/
final String[] mWords;
/*
////////////////////////////////////////////////
// Life-cycle
////////////////////////////////////////////////
*/
private WordResolver(String[] words, char[] index) {
mWords = words;
mData = index;
}
/**
* Tries to construct an instance given ordered set of words.
*
* Note: currently maximum number of words that can be contained
* is limited to {@link #MAX_WORDS}; additionally, maximum length
* of all such words can not exceed roughly 28000 characters.
*
* @return WordResolver constructed for given set of words, if
* the word set size is not too big; null to indicate "too big"
* instance.
*/
public static WordResolver constructInstance(TreeSet wordSet)
{
if (wordSet.size() > MAX_WORDS) {
return null;
}
return new Builder(wordSet).construct();
}
/*
////////////////////////////////////////////////
// Public API
////////////////////////////////////////////////
*/
/**
* @return Number of words contained
*/
public int size() {
return mWords.length;
}
/*
public int indexSize() {
return mData.length;
}
*/
/**
* @param str Character array that contains the word to find
* @param start Index of the first character of the word
* @param end Index following the last character of the word,
* so that end - start
equals word length (similar
* to the way String.substring()
has).
*
* @return (Shared) string instance of the word, if it exists in
* the word set; null if not.
*/
@SuppressWarnings("cast")
public String find(char[] str, final int start, final int end)
{
char[] data = mData;
// 03-Jan-2006, TSa: Special case; one entry
if (data == null) {
return findFromOne(str, start, end);
}
int ptr = 0; // pointer to compressed set data
int offset = start;
while (true) {
// End of input String? Need to match the runt entry!
if (offset == end) {
if (data[ptr+1] == CHAR_NULL) {
return mWords[data[ptr+2] - NEGATIVE_OFFSET];
}
return null;
}
int count = data[ptr++];
// Need to find the branch to follow, if any
char c = str[offset++];
inner_block:
do { // dummy loop, need to have break
// Linear or binary search?
if (count < MIN_BINARY_SEARCH) {
// always at least two branches; never less
if (data[ptr] == c) {
ptr = (int) data[ptr+1];
break inner_block;
}
if (data[ptr+2] == c) {
ptr = (int) data[ptr+3];
break inner_block;
}
int branchEnd = ptr + (count << 1);
// Starts from entry #3, if such exists
for (ptr += 4; ptr < branchEnd; ptr += 2) {
if (data[ptr] == c) {
ptr = (int) data[ptr+1];
break inner_block;
}
}
return null; // No match!
} else { // Ok, binary search:
int low = 0;
int high = count-1;
int mid;
while (low <= high) {
mid = (low + high) >> 1;
int ix = ptr + (mid << 1);
int diff = data[ix] - c;
if (diff > 0) { // char was 'higher', need to go down
high = mid-1;
} else if (diff < 0) { // lower, need to go up
low = mid+1;
} else { // match (so far)
ptr = (int) data[ix+1];
break inner_block;
}
}
return null; // No match!
}
} while (false);
// Ok; now, is it the end?
if (ptr >= NEGATIVE_OFFSET) {
String word = mWords[ptr - NEGATIVE_OFFSET];
int expLen = (end - start);
if (word.length() != expLen) {
return null;
}
for (int i = offset - start; offset < end; ++i, ++offset) {
if (word.charAt(i) != str[offset]) {
return null;
}
}
return word;
}
}
// never gets here
}
private String findFromOne(char[] str, final int start, final int end)
{
String word = mWords[0];
int len = end-start;
if (word.length() != len) {
return null;
}
for (int i = 0; i < len; ++i) {
if (word.charAt(i) != str[start+i]) {
return null;
}
}
return word;
}
/**
* @return (Shared) string instance of the word, if it exists in
* the word set; null if not.
*/
@SuppressWarnings("cast")
public String find(String str)
{
char[] data = mData;
// 03-Jan-2006, TSa: Special case; one entry
if (data == null) {
String word = mWords[0];
return word.equals(str) ? word : null;
}
int ptr = 0; // pointer to compressed set data
int offset = 0;
int end = str.length();
while (true) {
// End of input String? Need to match the runt entry!
if (offset == end) {
if (data[ptr+1] == CHAR_NULL) {
return mWords[data[ptr+2] - NEGATIVE_OFFSET];
}
return null;
}
int count = data[ptr++];
// Need to find the branch to follow, if any
char c = str.charAt(offset++);
inner_block:
do { // dummy loop, need to have break
// Linear or binary search?
if (count < MIN_BINARY_SEARCH) {
// always at least two branches; never less
if (data[ptr] == c) {
ptr = (int) data[ptr+1];
break inner_block;
}
if (data[ptr+2] == c) {
ptr = (int) data[ptr+3];
break inner_block;
}
int branchEnd = ptr + (count << 1);
// Starts from entry #3, if such exists
for (ptr += 4; ptr < branchEnd; ptr += 2) {
if (data[ptr] == c) {
ptr = (int) data[ptr+1];
break inner_block;
}
}
return null; // No match!
} else { // Ok, binary search:
int low = 0;
int high = count-1;
int mid;
while (low <= high) {
mid = (low + high) >> 1;
int ix = ptr + (mid << 1);
int diff = data[ix] - c;
if (diff > 0) { // char was 'higher', need to go down
high = mid-1;
} else if (diff < 0) { // lower, need to go up
low = mid+1;
} else { // match (so far)
ptr = (int) data[ix+1];
break inner_block;
}
}
return null; // No match!
}
} while (false);
// Ok; now, is it the end?
if (ptr >= NEGATIVE_OFFSET) {
String word = mWords[ptr - NEGATIVE_OFFSET];
if (word.length() != str.length()) {
return null;
}
for (; offset < end; ++offset) {
if (word.charAt(offset) != str.charAt(offset)) {
return null;
}
}
return word;
}
}
// never gets here
}
/*
////////////////////////////////////////////////
// Re-defined public methods
////////////////////////////////////////////////
*/
@Override
public String toString()
{
StringBuilder sb = new StringBuilder(16 + (mWords.length << 3));
for (int i = 0, len = mWords.length; i < len; ++i) {
if (i > 0) {
sb.append(", ");
}
sb.append(mWords[i]);
}
return sb.toString();
}
/*
////////////////////////////////////////////////
// Helper classes
////////////////////////////////////////////////
*/
private final static class Builder
{
final String[] mWords;
char[] mData;
/**
* Number of characters currently used from mData
*/
int mSize;
public Builder(TreeSet wordSet)
{
int wordCount = wordSet.size();
mWords = new String[wordCount];
wordSet.toArray(mWords);
/* 03-Jan-2006, TSa: Special case: just one entry; if so,
* let's leave char array null, and just have the String
* array with one entry.
*/
if (wordCount < 2) {
if (wordCount == 0) {
throw new IllegalArgumentException(); // not legal
}
mData = null;
} else {
/* Let's guess approximate size we should need, assuming
* average word length of 6 characters, overhead matching
* compression (ie. about 1-to-1 ratio overall)
*/
int size = wordCount * 6;
if (size < 256) {
size = 256;
}
mData = new char[size];
}
}
/**
* @return Raw character data that contains compressed structure
* of the word set
*/
public WordResolver construct()
{
char[] result;
/* 03-Jan-2006, TSa: Special case: just one entry; if so,
* let's leave char array null, and just have the String
* array with one entry.
*/
if (mData == null) {
result = null;
} else {
constructBranch(0, 0, mWords.length);
// Too big?
if (mSize > NEGATIVE_OFFSET) {
return null;
}
result = new char[mSize];
System.arraycopy(mData, 0, result, 0, mSize);
}
return new WordResolver(mWords, result);
}
/**
* Method that is called recursively to build the data
* representation for a branch, ie. part of word set tree
* that still has more than one ending
*
* @param charIndex Index of the character in words to consider
* for this round
* @param start Index of the first word to be processed
* @param end Index of the word after last word to be processed
* (so that number of words is end - start - 1
*/
@SuppressWarnings("cast")
private void constructBranch(int charIndex, int start, int end)
{
// If more than one entry, need to divide into groups
// First, need to add placeholder for branch count:
if (mSize >= mData.length) {
expand(1);
}
mData[mSize++] = 0; // placeholder!
/* structStart will point to second char of first entry
* (which will temporarily have entry count, eventually 'link'
* to continuation)
*/
int structStart = mSize + 1;
int groupCount = 0;
int groupStart = start;
String[] words = mWords;
boolean gotRunt;
/* First thing we need to do is a special check for the
* first entry -- it may be "runt" word, one that has no
* more chars but also has a longer version ("id" vs.
* "identifier"). If so, it needs to be marked; this is done
* by adding a special entry before other entries (since such
* entry would always be ordered first alphabetically)
*/
if (words[groupStart].length() == charIndex) { // yup, got one:
if ((mSize + 2) > mData.length) {
expand(2);
}
/* First null marks the "missing" char (or, end-of-word);
* and then we need the index
*/
mData[mSize++] = CHAR_NULL;
mData[mSize++] = (char) (NEGATIVE_OFFSET + groupStart);
// Ok, let's then ignore that entry
++groupStart;
++groupCount;
gotRunt = true;
} else {
gotRunt = false;
}
// Ok, then, let's find the ('real') groupings:
while (groupStart < end) {
// Inner loop, let's find the group:
char c = words[groupStart].charAt(charIndex);
int j = groupStart+1;
while (j < end && words[j].charAt(charIndex) == c) {
++j;
}
/* Ok, let's store the char in there, along with count;
* count will be needed in second, and will then get
* overwritten with actual data later on
*/
if ((mSize + 2) > mData.length) {
expand(2);
}
mData[mSize++] = c;
mData[mSize++] = (char) (j - groupStart); // entries in group
groupStart = j;
++groupCount;
}
/* Ok, groups found; need to loop through them, recursively
* calling branch and/or leaf methods
*/
// first let's output the header, ie. group count:
mData[structStart-2] = (char) groupCount;
groupStart = start;
// Do we have the "runt" to skip?
if (gotRunt) {
structStart += 2;
++groupStart;
}
int structEnd = mSize;
++charIndex;
for (; structStart < structEnd; structStart += 2) {
groupCount = (int) mData[structStart]; // no sign expansion, is ok
/* Ok, count gotten, can either create a branch (if more than
* one entry) or leaf (just one entry)
*/
if (groupCount == 1) {
mData[structStart] = (char) (NEGATIVE_OFFSET + groupStart);
} else {
mData[structStart] = (char) mSize;
constructBranch(charIndex, groupStart,
groupStart + groupCount);
}
groupStart += groupCount;
}
// done!
}
private char[] expand(int needSpace)
{
char[] old = mData;
int len = old.length;
int newSize = len + ((len < 4096) ? len : (len >> 1));
/* Let's verify we get enough; should always be true but
* better safe than sorry
*/
if (newSize < (mSize + needSpace)) {
newSize = mSize + needSpace + 64;
}
mData = new char[newSize];
System.arraycopy(old, 0, mData, 0, len);
return mData;
}
}
/*
////////////////////////////////////////////////////
// Simple test driver, useful for debugging
// (uncomment if needed -- commented out so it won't
// affect coverage testing)
////////////////////////////////////////////////////
*/
/*
public static void main(String[] args)
{
if (args.length < 2) {
System.err.println("Usage: "+WordResolver.class+" word1 [word2] ... [wordN] keyword");
System.exit(1);
}
String key = args[args.length-1];
TreeSet words = new TreeSet();
for (int i = 0; i < args.length-1; ++i) {
words.add(args[i]);
}
WordResolver set = WordResolver.constructInstance(words);
//outputData(set.mData);
// Ok, and then the test!
char[] keyA = new char[key.length() + 4];
key.getChars(0, key.length(), keyA, 2);
//System.out.println("Word '"+key+"' found via array search: "+WordResolver.find(data, keyA, 2, key.length() + 2));
System.out.println("Word '"+key+"' found via array search: "+set.find(keyA, 2, key.length() + 2));
}
static void outputData(char[] data)
{
for (int i = 0; i < data.length; ++i) {
char c = data[i];
System.out.print(Integer.toHexString(i)+" ["+Integer.toHexString(c)+"]");
if (c > 32 && c <= 127) { // printable char (letter)
System.out.println(" -> '"+c+"'");
} else {
System.out.println();
}
}
}
*/
}