我有一个Java程序，它存储了很多从Strings到各种对象的映射。
现在，我的选择是依赖哈希（通过HashMap）或二进制搜索（通过TreeMap）。我想知道在流行的高质量馆藏图书馆中是否有一个高效且标准的基于trie的地图实施？
我过去曾写过自己的作品，但如果有的话，我宁愿选择标准的东西。
快速澄清：虽然我的问题很普遍，但在当前项目中，我处理的是大量数据，这些数据由完全限定的类名或方法签名索引。因此，有许多共享前缀。
答案 0 :(得分：30)
您可能需要查看Google Guava的Trie implementation that Limewire is contributing。
答案 1 :(得分：9)
核心Java库中没有trie数据结构。
这可能是因为try通常用于存储字符串，而Java数据结构更通用，通常包含任何Object（定义相等和散列操作），尽管它们有时仅限于{{1对象（定义顺序）。尽管Comparable适用于字符串，但我认为你可以对CharSequence做其他类型的符号。
这是另一个要考虑的问题：当尝试在Java中实现传统的trie时，很快就会面临Java支持Unicode的事实。要获得任何类型的空间效率，必须将trie中的字符串限制为某些符号子集，或者放弃将子节点存储在由符号索引的数组中的传统方法。这可能是为什么尝试被认为不足以包含在核心库中的另一个原因，以及如果您实现自己的或使用第三方库需要注意的事项。
答案 2 :(得分：7)
同时查看concurrent-trees。它们支持Radix和Suffix树，专为高并发环境而设计。
答案 3 :(得分：4)
Apache Commons Collections v4.0现在支持trie结构。
有关详细信息，请参阅org.apache.commons.collections4.trie package info。特别要检查PatriciaTrie类：
实施PATRICIA Trie（用于检索以字母数字编码的信息的实用算法）。
PATRICIA Trie是一个压缩的Trie。 PATRICIA不是将所有数据存储在Trie的边缘（并且具有空的内部节点），而是将数据存储在每个节点中。这允许非常有效的遍历，插入，删除，前驱，后继，前缀，范围和选择（对象）操作。所有操作在O（K）时间内最差地执行，其中K是树中最大项目中的位数。实际上，操作实际上需要O（A（K））时间，其中A（K）是树中所有项目的平均位数。
答案 4 :(得分：3)
我编写并发布了一个简单快速的实现here。
答案 5 :(得分：2)
Apache的公共收藏： org.apache.commons.collections4.trie.PatriciaTrie
答案 6 :(得分：1)
我认为你需要的是org.apache.commons.collections.FastTreeMap。
答案 7 :(得分：1)
您可以尝试使用Completely Java库，它具有PatriciaTrie实现功能。 API很小且易于上手，并且可以在Maven central repository中找到。
答案 8 :(得分：0)
您也可以查看this TopCoder一个（需要注册......）。
答案 9 :(得分：0)
如果您需要排序地图，那么尝试是值得的。 如果不这样，那么hashmap会更好。 与标准Java实现相比，可以改进带字符串键的Hashmap： Array hash map
答案 10 :(得分：0)
如果您不担心拉入Scala库，可以使用我写的burst trie这个节省空间的实现。
答案 11 :(得分：0)
这是我的实现，通过以下方式享受：GitHub - MyTrie.java
/* usage:
MyTrie trie = new MyTrie();
trie.insert("abcde");
trie.insert("abc");
trie.insert("sadas");
trie.insert("abc");
trie.insert("wqwqd");
System.out.println(trie.contains("abc"));
System.out.println(trie.contains("abcd"));
System.out.println(trie.contains("abcdefg"));
System.out.println(trie.contains("ab"));
System.out.println(trie.getWordCount("abc"));
System.out.println(trie.getAllDistinctWords());
*/
import java.util.*;
public class MyTrie {
private class Node {
public int[] next = new int[26];
public int wordCount;
public Node() {
for(int i=0;i<26;i++) {
next[i] = NULL;
}
wordCount = 0;
}
}
private int curr;
private Node[] nodes;
private List<String> allDistinctWords;
public final static int NULL = -1;
public MyTrie() {
nodes = new Node[100000];
nodes[0] = new Node();
curr = 1;
}
private int getIndex(char c) {
return (int)(c - 'a');
}
private void depthSearchWord(int x, String currWord) {
for(int i=0;i<26;i++) {
int p = nodes[x].next[i];
if(p != NULL) {
String word = currWord + (char)(i + 'a');
if(nodes[p].wordCount > 0) {
allDistinctWords.add(word);
}
depthSearchWord(p, word);
}
}
}
public List<String> getAllDistinctWords() {
allDistinctWords = new ArrayList<String>();
depthSearchWord(0, "");
return allDistinctWords;
}
public int getWordCount(String str) {
int len = str.length();
int p = 0;
for(int i=0;i<len;i++) {
int j = getIndex(str.charAt(i));
if(nodes[p].next[j] == NULL) {
return 0;
}
p = nodes[p].next[j];
}
return nodes[p].wordCount;
}
public boolean contains(String str) {
int len = str.length();
int p = 0;
for(int i=0;i<len;i++) {
int j = getIndex(str.charAt(i));
if(nodes[p].next[j] == NULL) {
return false;
}
p = nodes[p].next[j];
}
return nodes[p].wordCount > 0;
}
public void insert(String str) {
int len = str.length();
int p = 0;
for(int i=0;i<len;i++) {
int j = getIndex(str.charAt(i));
if(nodes[p].next[j] == NULL) {
nodes[curr] = new Node();
nodes[p].next[j] = curr;
curr++;
}
p = nodes[p].next[j];
}
nodes[p].wordCount++;
}
}
答案 12 :(得分：0)
下面是Trie的基本HashMap实现。有些人可能觉得这很有用......
class Trie {
HashMap<Character, HashMap> root;
public Trie() {
root = new HashMap<Character, HashMap>();
}
public void addWord(String word) {
HashMap<Character, HashMap> node = root;
for (int i = 0; i < word.length(); i++) {
Character currentLetter = word.charAt(i);
if (node.containsKey(currentLetter) == false) {
node.put(currentLetter, new HashMap<Character, HashMap>());
}
node = node.get(currentLetter);
}
}
public boolean containsPrefix(String word) {
HashMap<Character, HashMap> node = root;
for (int i = 0; i < word.length(); i++) {
Character currentLetter = word.charAt(i);
if (node.containsKey(currentLetter)) {
node = node.get(currentLetter);
} else {
return false;
}
}
return true;
}
}
答案 13 :(得分：0)
我刚刚尝试了自己的Concurrent TRIE实现，但没有基于字符，它基于HashCode。我们仍然可以使用每个CHAR hascode的Map Map 您可以使用代码@ https://github.com/skanagavelu/TrieHashMap/blob/master/src/TrieMapPerformanceTest.java对此进行测试 https://github.com/skanagavelu/TrieHashMap/blob/master/src/TrieMapValidationTest.java
import java.util.concurrent.atomic.AtomicReferenceArray;
public class TrieMap {
public static int SIZEOFEDGE = 4;
public static int OSIZE = 5000;
}
abstract class Node {
public Node getLink(String key, int hash, int level){
throw new UnsupportedOperationException();
}
public Node createLink(int hash, int level, String key, String val) {
throw new UnsupportedOperationException();
}
public Node removeLink(String key, int hash, int level){
throw new UnsupportedOperationException();
}
}
class Vertex extends Node {
String key;
volatile String val;
volatile Vertex next;
public Vertex(String key, String val) {
this.key = key;
this.val = val;
}
@Override
public boolean equals(Object obj) {
Vertex v = (Vertex) obj;
return this.key.equals(v.key);
}
@Override
public int hashCode() {
return key.hashCode();
}
@Override
public String toString() {
return key +"@"+key.hashCode();
}
}
class Edge extends Node {
volatile AtomicReferenceArray<Node> array; //This is needed to ensure array elements are volatile
public Edge(int size) {
array = new AtomicReferenceArray<Node>(8);
}
@Override
public Node getLink(String key, int hash, int level){
int index = Base10ToBaseX.getBaseXValueOnAtLevel(Base10ToBaseX.Base.BASE8, hash, level);
Node returnVal = array.get(index);
for(;;) {
if(returnVal == null) {
return null;
}
else if((returnVal instanceof Vertex)) {
Vertex node = (Vertex) returnVal;
for(;node != null; node = node.next) {
if(node.key.equals(key)) {
return node;
}
}
return null;
} else { //instanceof Edge
level = level + 1;
index = Base10ToBaseX.getBaseXValueOnAtLevel(Base10ToBaseX.Base.BASE8, hash, level);
Edge e = (Edge) returnVal;
returnVal = e.array.get(index);
}
}
}
@Override
public Node createLink(int hash, int level, String key, String val) { //Remove size
for(;;) { //Repeat the work on the current node, since some other thread modified this node
int index = Base10ToBaseX.getBaseXValueOnAtLevel(Base10ToBaseX.Base.BASE8, hash, level);
Node nodeAtIndex = array.get(index);
if ( nodeAtIndex == null) {
Vertex newV = new Vertex(key, val);
boolean result = array.compareAndSet(index, null, newV);
if(result == Boolean.TRUE) {
return newV;
}
//continue; since new node is inserted by other thread, hence repeat it.
}
else if(nodeAtIndex instanceof Vertex) {
Vertex vrtexAtIndex = (Vertex) nodeAtIndex;
int newIndex = Base10ToBaseX.getBaseXValueOnAtLevel(Base10ToBaseX.Base.BASE8, vrtexAtIndex.hashCode(), level+1);
int newIndex1 = Base10ToBaseX.getBaseXValueOnAtLevel(Base10ToBaseX.Base.BASE8, hash, level+1);
Edge edge = new Edge(Base10ToBaseX.Base.BASE8.getLevelZeroMask()+1);
if(newIndex != newIndex1) {
Vertex newV = new Vertex(key, val);
edge.array.set(newIndex, vrtexAtIndex);
edge.array.set(newIndex1, newV);
boolean result = array.compareAndSet(index, vrtexAtIndex, edge); //REPLACE vertex to edge
if(result == Boolean.TRUE) {
return newV;
}
//continue; since vrtexAtIndex may be removed or changed to Edge already.
} else if(vrtexAtIndex.key.hashCode() == hash) {//vrtex.hash == hash) { HERE newIndex == newIndex1
synchronized (vrtexAtIndex) {
boolean result = array.compareAndSet(index, vrtexAtIndex, vrtexAtIndex); //Double check this vertex is not removed.
if(result == Boolean.TRUE) {
Vertex prevV = vrtexAtIndex;
for(;vrtexAtIndex != null; vrtexAtIndex = vrtexAtIndex.next) {
prevV = vrtexAtIndex; // prevV is used to handle when vrtexAtIndex reached NULL
if(vrtexAtIndex.key.equals(key)){
vrtexAtIndex.val = val;
return vrtexAtIndex;
}
}
Vertex newV = new Vertex(key, val);
prevV.next = newV; // Within SYNCHRONIZATION since prevV.next may be added with some other.
return newV;
}
//Continue; vrtexAtIndex got changed
}
} else { //HERE newIndex == newIndex1 BUT vrtex.hash != hash
edge.array.set(newIndex, vrtexAtIndex);
boolean result = array.compareAndSet(index, vrtexAtIndex, edge); //REPLACE vertex to edge
if(result == Boolean.TRUE) {
return edge.createLink(hash, (level + 1), key, val);
}
}
}
else { //instanceof Edge
return nodeAtIndex.createLink(hash, (level + 1), key, val);
}
}
}
@Override
public Node removeLink(String key, int hash, int level){
for(;;) {
int index = Base10ToBaseX.getBaseXValueOnAtLevel(Base10ToBaseX.Base.BASE8, hash, level);
Node returnVal = array.get(index);
if(returnVal == null) {
return null;
}
else if((returnVal instanceof Vertex)) {
synchronized (returnVal) {
Vertex node = (Vertex) returnVal;
if(node.next == null) {
if(node.key.equals(key)) {
boolean result = array.compareAndSet(index, node, null);
if(result == Boolean.TRUE) {
return node;
}
continue; //Vertex may be changed to Edge
}
return null; //Nothing found; This is not the same vertex we are looking for. Here hashcode is same but key is different.
} else {
if(node.key.equals(key)) { //Removing the first node in the link
boolean result = array.compareAndSet(index, node, node.next);
if(result == Boolean.TRUE) {
return node;
}
continue; //Vertex(node) may be changed to Edge, so try again.
}
Vertex prevV = node; // prevV is used to handle when vrtexAtIndex is found and to be removed from its previous
node = node.next;
for(;node != null; prevV = node, node = node.next) {
if(node.key.equals(key)) {
prevV.next = node.next; //Removing other than first node in the link
return node;
}
}
return null; //Nothing found in the linked list.
}
}
} else { //instanceof Edge
return returnVal.removeLink(key, hash, (level + 1));
}
}
}
}
class Base10ToBaseX {
public static enum Base {
/**
* Integer is represented in 32 bit in 32 bit machine.
* There we can split this integer no of bits into multiples of 1,2,4,8,16 bits
*/
BASE2(1,1,32), BASE4(3,2,16), BASE8(7,3,11)/* OCTAL*/, /*BASE10(3,2),*/
BASE16(15, 4, 8){
public String getFormattedValue(int val){
switch(val) {
case 10:
return "A";
case 11:
return "B";
case 12:
return "C";
case 13:
return "D";
case 14:
return "E";
case 15:
return "F";
default:
return "" + val;
}
}
}, /*BASE32(31,5,1),*/ BASE256(255, 8, 4), /*BASE512(511,9),*/ Base65536(65535, 16, 2);
private int LEVEL_0_MASK;
private int LEVEL_1_ROTATION;
private int MAX_ROTATION;
Base(int levelZeroMask, int levelOneRotation, int maxPossibleRotation) {
this.LEVEL_0_MASK = levelZeroMask;
this.LEVEL_1_ROTATION = levelOneRotation;
this.MAX_ROTATION = maxPossibleRotation;
}
int getLevelZeroMask(){
return LEVEL_0_MASK;
}
int getLevelOneRotation(){
return LEVEL_1_ROTATION;
}
int getMaxRotation(){
return MAX_ROTATION;
}
String getFormattedValue(int val){
return "" + val;
}
}
public static int getBaseXValueOnAtLevel(Base base, int on, int level) {
if(level > base.getMaxRotation() || level < 1) {
return 0; //INVALID Input
}
int rotation = base.getLevelOneRotation();
int mask = base.getLevelZeroMask();
if(level > 1) {
rotation = (level-1) * rotation;
mask = mask << rotation;
} else {
rotation = 0;
}
return (on & mask) >>> rotation;
}
}