字典树Trie

目录

什么是字典树

 基于TreeMap的字典树

 不使用map实现字典树

 例题208实现一个前缀树

练习211添加和搜索单词

---

 

• 什么是字典树

 

•  基于TreeMap的字典树

package com.suanfa.trie;

import java.util.TreeMap;

/**
 * 字典树（存储单词）基于TreeMap
 * @author Administrator
 *
 */
public class Trie {
	private class Node{

        public boolean isWord;//是否是一个单词
        public TreeMap next;//存储字母

        public Node(boolean isWord){
            this.isWord = isWord;
            next = new TreeMap();
        }

        public Node(){
            this(false);
        }
    }

    private Node root;//根节点
    private int size;

    public Trie(){
        root = new Node();
        size = 0;
    }

    // 获得Trie中存储的单词数量
    public int getSize(){
        return size;
    }

    // 向Trie中添加一个新的单词word
    public void add(String word){

        Node cur = root;//从根节点开始添加
        for(int i = 0 ; i < word.length() ; i ++){
            char c = word.charAt(i);//获取单词中的每一个字母
            if(cur.next.get(c) == null)//当前节点的下一个节点中没有存储这个字母，添加进去
                cur.next.put(c, new Node());
            cur = cur.next.get(c);//否则向下寻找
        }
        //如果不是一个单词则为空格，单词数加一
        if(!cur.isWord){
            cur.isWord = true;
            size ++;
        }
    }

    // 查询单词word是否在Trie中
    public boolean contains(String word){
    	//从根节点开始遍历
        Node cur = root;
        for(int i = 0 ; i < word.length() ; i ++){
            char c = word.charAt(i);
            if(cur.next.get(c) == null)
                return false;
            cur = cur.next.get(c);
        }
        return cur.isWord;
    }

    // 查询是否在Trie中有单词以prefix为前缀
    public boolean isPrefix(String prefix){

        Node cur = root;
        for(int i = 0 ; i < prefix.length() ; i ++){
            char c = prefix.charAt(i);
            if(cur.next.get(c) == null)
                return false;
            cur = cur.next.get(c);
        }

        return true;
    }

}

•  不使用map实现字典树

package com.suanfa.trie;
/**
 * 字典树，不用map
 * @author Administrator
 *
 */
public class Trie3 {

    private class Node{

        public boolean isWord;
        public Node[] next;

        public Node(boolean isWord){
            this.isWord = isWord;
            next = new Node[26];
        }

        public Node(){
            this(false);
        }
    }

    private Node root;
    private int size;

    public Trie3(){
        root = new Node();
        size = 0;
    }

    // 获得Trie中存储的单词数量
    public int getSize(){
        return size;
    }

    // 向Trie中添加一个新的单词word
    public void add(String word){

        Node cur = root;
        for(int i = 0 ; i < word.length() ; i ++){
            char c = word.charAt(i);
            if(cur.next[c-'a'] == null)
                cur.next[c-'a'] = new Node();
            cur = cur.next[c-'a'];
        }

        if(!cur.isWord){
            cur.isWord = true;
            size ++;
        }
    }

    // 查询单词word是否在Trie中
    public boolean contains(String word){

        Node cur = root;
        for(int i = 0 ; i < word.length() ; i ++){
            char c = word.charAt(i);
            if(cur.next[c-'a'] == null)
                return false;
            cur = cur.next[c-'a'];
        }
        return cur.isWord;
    }
}

•  例题208实现一个前缀树

package com.suanfa.trie;

/// 208. Implement Trie (Prefix Tree)
/// https://leetcode.com/problems/implement-trie-prefix-tree/description/

import java.util.TreeMap;
/**
 * 实现一个前缀树
 * @author Administrator
 *
 */
public class Trie208 {

    private class Node{

        public boolean isWord;
        public TreeMap next;

        public Node(boolean isWord){
            this.isWord = isWord;
            next = new TreeMap();
        }

        public Node(){
            this(false);
        }
    }

    private Node root;

    public Trie208(){
        root = new Node();
    }

    // 向Trie中添加一个新的单词word
    public void insert(String word){

        Node cur = root;
        for(int i = 0 ; i < word.length() ; i ++){
            char c = word.charAt(i);
            if(cur.next.get(c) == null)
                cur.next.put(c, new Node());
            cur = cur.next.get(c);
        }
        cur.isWord = true;
    }

    // 查询单词word是否在Trie中
    public boolean search(String word){

        Node cur = root;
        for(int i = 0 ; i < word.length() ; i ++){
            char c = word.charAt(i);
            if(cur.next.get(c) == null)
                return false;
            cur = cur.next.get(c);
        }
        return cur.isWord;
    }

    // 查询是否在Trie中有单词以prefix为前缀
    public boolean startsWith(String isPrefix){

        Node cur = root;
        for(int i = 0 ; i < isPrefix.length() ; i ++){
            char c = isPrefix.charAt(i);
            if(cur.next.get(c) == null)
                return false;
            cur = cur.next.get(c);
        }

        return true;
    }
}

• 练习211添加和搜索单词

 

package com.suanfa.trie;
/// Leetcode 211. Add and Search Word - Data structure design
/// https://leetcode.com/problems/add-and-search-word-data-structure-design/description/

import java.util.TreeMap;
/**
 * 添加和搜索单词
 * @author Administrator
 *
 */
public class WordDictionary {

    private class Node{

        public boolean isWord;
        public TreeMap next;

        public Node(boolean isWord){
            this.isWord = isWord;
            next = new TreeMap();
        }

        public Node(){
            this(false);
        }
    }

    private Node root;

    /** Initialize your data structure here. */
    public WordDictionary() {
        root = new Node();
    }

    /**添加单词 Adds a word into the data structure. */
    public void addWord(String word) {

        Node cur = root;
        for(int i = 0 ; i < word.length() ; i ++){
            char c = word.charAt(i);
            if(cur.next.get(c) == null)
                cur.next.put(c, new Node());
            cur = cur.next.get(c);
        }
        cur.isWord = true;
    }

    /**.可以代表任何字母， Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    public boolean search(String word) {
        return match(root, word, 0);
    }

    private boolean match(Node node, String word, int index){

        if(index == word.length())
            return node.isWord;

        char c = word.charAt(index);

        if(c != '.'){
            if(node.next.get(c) == null)
                return false;
            return match(node.next.get(c), word, index + 1);
        }
        else{
            for(char nextChar: node.next.keySet())
                if(match(node.next.get(nextChar), word, index + 1))
                    return true;
            return false;
        }
    }
}