玩转数据结构之字典树（前缀树 Trie）

Trie Basics

import java.util.TreeMap;

public class Trie {

    private class Node{

        public boolean isWord;
        public TreeMap<Character, Node> next;

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

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

    private Node root;
    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;
        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);
        }

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

searchig in trie

    // 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.get(c) == null)
                return false;
            cur = cur.next.get(c);
        }
        return cur.isWord;
    }

前缀查询

   // 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;
    }

Leetcode208 实现Trie

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

import java.util.TreeMap;
public class Trie208 {

    private class Node{

        public boolean isWord;
        public TreeMap<Character, Node> 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;
    }
}

Leetcode211 添加与搜索单词

/// 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;

public class WordDictionary {

    private class Node{

        public boolean isWord;
        public TreeMap<Character, Node> 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();
    }

    /** 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;
    }

    /** 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;
        }
    }
}

Leetcode 677 Map Sum Pairs

import java.util.TreeMap;

public class MapSum {

    private class Node{

        public int value;
        public TreeMap<Character, Node> next;

        public Node(int value){
            this.value = value;
            next = new TreeMap<>();
        }

        public Node(){
            this(0);
        }
    }

    private Node root;

    /** Initialize your data structure here. */
    public MapSum() {

        root = new Node();
    }

    public void insert(String key, int val) {

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

    public int sum(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 0;
            cur = cur.next.get(c);
        }

        return sum(cur);
    }

    private int sum(Node node){

        int res = node.value;
// 遍历所有的子树，把value值加入res
        for(char c: node.next.keySet())
            res += sum(node.next.get(c));
        return res;
    }
}

Trie的删除操作

• 当搜索到该单词的最后一个字母时，自底向上的删除
• 当首字母被公用的时候，保留首字母，只删除后面的字母就行
• 当删除的最后一个字母根本不是叶子节点的时候，直接删除isword就行

Trie局限性