Java双数组Trie树的实现方案总结
jopen
10年前
传统的Trie实现简单，但是占用的空间实在是难以接受，特别是当字符集不仅限于英文26个字符的时候，爆炸起来的空间根本无法接受。
双数组Trie就是优化了空间的Trie树，原理本文就不讲了，请参考An Efficient Implementation of Trie Structures，本程序的编写也是参考这篇论文的。
关于几点论文没有提及的细节和与论文不一一致的实现：
1.对于插入字符串，如果有一个字符串是另一个字符串的子串的话，我是将结束符也作为一条边，产生一个新的结点，这个结点新节点的Base我置为0
所以一个字符串结束也有2中情况：一个是Base值为负，存储剩余字符(可能只有一个结束符)到Tail数组；另一个是Base为0。
所以在查询的时候要考虑一下这两种情况
2.对于第一种冲突（论文中的Case 3），可能要将Tail中的字符串取出一部分，作为边放到索引中。论文是使用将尾串左移的方式，我的方式直接修改Base值，而不是移动尾串。
下面是java实现的代码，可以处理相同字符串插入，子串的插入等情况
/* * Name: Double Array Trie * Author: Yaguang Ding * Mail: dingyaguang117@gmail.com * Blog: blog.csdn.net/dingyaguang117 * Date: 2012/5/21 * Note: a word ends may be either of these two case: * 1. Base[cur_p] == pos ( pos<0 and Tail[-pos] == 'END_CHAR' ) * 2. Check[Base[cur_p] + Code('END_CHAR')] == cur_p */ import java.util.ArrayList; import java.util.HashMap; import java.util.Map; import java.util.Arrays; public class DoubleArrayTrie { final char END_CHAR = '\0'; final int DEFAULT_LEN = 1024; int Base[] = new int [DEFAULT_LEN]; int Check[] = new int [DEFAULT_LEN]; char Tail[] = new char [DEFAULT_LEN]; int Pos = 1; Map<Character ,Integer> CharMap = new HashMap<Character,Integer>(); ArrayList<Character> CharList = new ArrayList<Character>(); public DoubleArrayTrie() { Base[1] = 1; CharMap.put(END_CHAR,1); CharList.add(END_CHAR); CharList.add(END_CHAR); for(int i=0;i<26;++i) { CharMap.put((char)('a'+i),CharMap.size()+1); CharList.add((char)('a'+i)); } } private void Extend_Array() { Base = Arrays.copyOf(Base, Base.length*2); Check = Arrays.copyOf(Check, Check.length*2); } private void Extend_Tail() { Tail = Arrays.copyOf(Tail, Tail.length*2); } private int GetCharCode(char c) { if (!CharMap.containsKey(c)) { CharMap.put(c,CharMap.size()+1); CharList.add(c); } return CharMap.get(c); } private int CopyToTailArray(String s,int p) { int _Pos = Pos; while(s.length()-p+1 > Tail.length-Pos) { Extend_Tail(); } for(int i=p; i<s.length();++i) { Tail[_Pos] = s.charAt(i); _Pos++; } return _Pos; } private int x_check(Integer []set) { for(int i=1; ; ++i) { boolean flag = true; for(int j=0;j<set.length;++j) { int cur_p = i+set[j]; if(cur_p>= Base.length) Extend_Array(); if(Base[cur_p]!= 0 || Check[cur_p]!= 0) { flag = false; break; } } if (flag) return i; } } private ArrayList<Integer> GetChildList(int p) { ArrayList<Integer> ret = new ArrayList<Integer>(); for(int i=1; i<=CharMap.size();++i) { if(Base[p]+i >= Check.length) break; if(Check[Base[p]+i] == p) { ret.add(i); } } return ret; } private boolean TailContainString(int start,String s2) { for(int i=0;i<s2.length();++i) { if(s2.charAt(i) != Tail[i+start]) return false; } return true; } private boolean TailMatchString(int start,String s2) { s2 += END_CHAR; for(int i=0;i<s2.length();++i) { if(s2.charAt(i) != Tail[i+start]) return false; } return true; } public void Insert(String s) throws Exception { s += END_CHAR; int pre_p = 1; int cur_p; for(int i=0; i<s.length(); ++i) { //获取状态位置 cur_p = Base[pre_p]+GetCharCode(s.charAt(i)); //如果长度超过现有，拓展数组 if (cur_p >= Base.length) Extend_Array(); //空闲状态 if(Base[cur_p] == 0 && Check[cur_p] == 0) { Base[cur_p] = -Pos; Check[cur_p] = pre_p; Pos = CopyToTailArray(s,i+1); break; }else //已存在状态 if(Base[cur_p] > 0 && Check[cur_p] == pre_p) { pre_p = cur_p; continue; }else //冲突 1：遇到 Base[cur_p]小于0的，即遇到一个被压缩存到Tail中的字符串 if(Base[cur_p] < 0 && Check[cur_p] == pre_p) { int head = -Base[cur_p]; if(s.charAt(i+1)== END_CHAR && Tail[head]==END_CHAR) //插入重复字符串 { break; } //公共字母的情况，因为上一个判断已经排除了结束符，所以一定是2个都不是结束符 if (Tail[head] == s.charAt(i+1)) { int avail_base = x_check(new Integer[]{GetCharCode(s.charAt(i+1))}); Base[cur_p] = avail_base; Check[avail_base+GetCharCode(s.charAt(i+1))] = cur_p; Base[avail_base+GetCharCode(s.charAt(i+1))] = -(head+1); pre_p = cur_p; continue; } else { //2个字母不相同的情况，可能有一个为结束符 int avail_base ; avail_base = x_check(new Integer[]{GetCharCode(s.charAt(i+1)),GetCharCode(Tail[head])}); Base[cur_p] = avail_base; Check[avail_base+GetCharCode(Tail[head])] = cur_p; Check[avail_base+GetCharCode(s.charAt(i+1))] = cur_p; //Tail 为END_FLAG 的情况 if(Tail[head] == END_CHAR) Base[avail_base+GetCharCode(Tail[head])] = 0; else Base[avail_base+GetCharCode(Tail[head])] = -(head+1); if(s.charAt(i+1) == END_CHAR) Base[avail_base+GetCharCode(s.charAt(i+1))] = 0; else Base[avail_base+GetCharCode(s.charAt(i+1))] = -Pos; Pos = CopyToTailArray(s,i+2); break; } }else //冲突2：当前结点已经被占用，需要调整pre的base if(Check[cur_p] != pre_p) { ArrayList<Integer> list1 = GetChildList(pre_p); int toBeAdjust; ArrayList<Integer> list = null; if(true) { toBeAdjust = pre_p; list = list1; } int origin_base = Base[toBeAdjust]; list.add(GetCharCode(s.charAt(i))); int avail_base = x_check((Integer[])list.toArray(new Integer[list.size()])); list.remove(list.size()-1); Base[toBeAdjust] = avail_base; for(int j=0; j<list.size(); ++j) { //BUG int tmp1 = origin_base + list.get(j); int tmp2 = avail_base + list.get(j); Base[tmp2] = Base[tmp1]; Check[tmp2] = Check[tmp1]; //有后续 if(Base[tmp1] > 0) { ArrayList<Integer> subsequence = GetChildList(tmp1); for(int k=0; k<subsequence.size(); ++k) { Check[Base[tmp1]+subsequence.get(k)] = tmp2; } } Base[tmp1] = 0; Check[tmp1] = 0; } //更新新的cur_p cur_p = Base[pre_p]+GetCharCode(s.charAt(i)); if(s.charAt(i) == END_CHAR) Base[cur_p] = 0; else Base[cur_p] = -Pos; Check[cur_p] = pre_p; Pos = CopyToTailArray(s,i+1); break; } } } public boolean Exists(String word) { int pre_p = 1; int cur_p = 0; for(int i=0;i<word.length();++i) { cur_p = Base[pre_p]+GetCharCode(word.charAt(i)); if(Check[cur_p] != pre_p) return false; if(Base[cur_p] < 0) { if(TailMatchString(-Base[cur_p],word.substring(i+1))) return true; return false; } pre_p = cur_p; } if(Check[Base[cur_p]+GetCharCode(END_CHAR)] == cur_p) return true; return false; } //内部函数，返回匹配单词的最靠后的Base index， class FindStruct { int p; String prefix=""; } private FindStruct Find(String word) { int pre_p = 1; int cur_p = 0; FindStruct fs = new FindStruct(); for(int i=0;i<word.length();++i) { // BUG fs.prefix += word.charAt(i); cur_p = Base[pre_p]+GetCharCode(word.charAt(i)); if(Check[cur_p] != pre_p) { fs.p = -1; return fs; } if(Base[cur_p] < 0) { if(TailContainString(-Base[cur_p],word.substring(i+1))) { fs.p = cur_p; return fs; } fs.p = -1; return fs; } pre_p = cur_p; } fs.p = cur_p; return fs; } public ArrayList<String> GetAllChildWord(int index) { ArrayList<String> result = new ArrayList<String>(); if(Base[index] == 0) { result.add(""); return result; } if(Base[index] < 0) { String r=""; for(int i=-Base[index];Tail[i]!=END_CHAR;++i) { r+= Tail[i]; } result.add(r); return result; } for(int i=1;i<=CharMap.size();++i) { if(Check[Base[index]+i] == index) { for(String s:GetAllChildWord(Base[index]+i)) { result.add(CharList.get(i)+s); } //result.addAll(GetAllChildWord(Base[index]+i)); } } return result; } public ArrayList<String> FindAllWords(String word) { ArrayList<String> result = new ArrayList<String>(); String prefix = ""; FindStruct fs = Find(word); int p = fs.p; if (p == -1) return result; if(Base[p]<0) { String r=""; for(int i=-Base[p];Tail[i]!=END_CHAR;++i) { r+= Tail[i]; } result.add(fs.prefix+r); return result; } if(Base[p] > 0) { ArrayList<String> r = GetAllChildWord(p); for(int i=0;i<r.size();++i) { r.set(i, fs.prefix+r.get(i)); } return r; } return result; } } 测试代码：
import java.io.BufferedReader; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Scanner; import javax.xml.crypto.Data; public class Main { public static void main(String[] args) throws Exception { ArrayList<String> words = new ArrayList<String>(); BufferedReader reader = new BufferedReader(new InputStreamReader(new FileInputStream("E:/兔子的试验学习中心[课内]/ACM大赛/ACM第四届校赛/E命令提示/words3.dic"))); String s; int num = 0; while((s=reader.readLine()) != null) { words.add(s); num ++; } DoubleArrayTrie dat = new DoubleArrayTrie(); for(String word: words) { dat.Insert(word); } System.out.println(dat.Base.length); System.out.println(dat.Tail.length); Scanner sc = new Scanner(System.in); while(sc.hasNext()) { String word = sc.next(); System.out.println(dat.Exists(word)); System.out.println(dat.FindAllWords(word)); } } } 下面是测试结果，构造6W英文单词的DAT，大概需要20秒
我增长数组的时候是每次长度增加到2倍，初始1024
Base和Check数组的长度为131072
Tail的长度为262144
