com.mayabot.nlp.collection.dat.DATLongMatcher Maven / Gradle / Ivy
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/*
* Copyright 2018 mayabot.com authors. All rights reserved.
*
* 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.
*/
/**
* DoubleArrayTrieMap: Java implementation of Darts (Double-ARray Trie System)
*
*
* Copyright(C) 2001-2007 Taku Kudo <[email protected]>
* Copyright(C) 2009 MURAWAKI Yugo <[email protected]>
* Copyright(C) 2012 KOMIYA Atsushi <[email protected]>
*
*
*
* The contents of this file may be used under the terms of either of the GNU
* Lesser General Public License Version 2.1 or later (the "LGPL"), or the BSD
* License (the "BSD").
*
*/
/*
* 源代码参考和部分引用来自 https://github.com/hankcs/HanLP https://github.com/NLPchina/ansj_seg
*/
package com.mayabot.nlp.collection.dat;
/**
* 一个搜索工具（注意，当调用next()返回false后不应该继续调用next()，除非reset状态）
*
* DAT的匹配器是一个多匹配器，把各种可能都计算出来
*/
public class DATLongMatcher {
/**
* key的起点(在原始文本中)
*/
private int begin;
/**
* key的长度
*/
private int length;
/**
* key的字典序坐标
*/
private int index;
/**
* 传入的字符数组
*/
private char[] charArray;
/**
* 上一个node位置
*/
private int last;
/**
* 上一个字符的下标
*/
private int i;
/**
* charArray的长度，效率起见，开个变量
*/
private int arrayLength;
private DoubleArrayTrie dat;
private int[] base;
private int[] check;
private Object value = null;
private Object obj = new Object();
/**
* 构造一个双数组搜索工具
*
* @param offset 搜索的起始位置
* @param text 搜索的目标字符数组
*/
DATLongMatcher(DoubleArrayTrie dat, String text, int offset) {
this(dat, text.toCharArray(), offset);
}
DATLongMatcher(DoubleArrayTrie dat, char[] charArray, int offset) {
this.charArray = charArray;
this.dat = dat;
i = offset;
last = dat.base[0];
arrayLength = charArray.length;
// A trick，如果文本长度为0的话，调用next()时，会带来越界的问题。
// 所以我要在第一次调用next()的时候触发begin == arrayLength进而返回false。
// 当然也可以改成begin >= arrayLength，不过我觉得操作符>=的效率低于==
if (arrayLength == 0) {
begin = -1;
} else {
begin = offset;
}
base = dat.base;
check = dat.check;
}
/**
* 取出下一个命中输出
*
* @return 是否命中，当返回false表示搜索结束，否则使用公开的成员读取命中的详细信息
*/
public boolean next() {
value = null;
begin = i;
int b = base[0];
int n;
int p;
for (; ; ++i) {
if (i >= arrayLength) // 指针到头了，将起点往前挪一个，重新开始，状态归零
{
return value != null;
}
p = b + (int) (charArray[i]) + 1; // 状态转移 p = base[char[i-1]] + char[i] + 1
if (b == check[p]) // base[char[i-1]] == check[base[char[i-1]] + char[i] + 1]
{
b = base[p]; // 转移成功
} else {
if (begin == arrayLength) {
break;
}
if (value != null) {
i = begin + length; // 输出最长词后，从该词语的下一个位置恢复扫描
return true;
}
i = begin; // 转移失败，也将起点往前挪一个，重新开始，状态归零
++begin;
b = base[0];
}
p = b;
n = base[p];
if (b == check[p] && n < 0) // base[p] == check[p] && base[p] < 0 查到一个词
{
length = i - begin + 1;
index = -n - 1;
value = obj;
}
}
return false;
}
public int getBegin() {
return begin;
}
public int getLength() {
return length;
}
public int getIndex() {
return index;
}
}