为了账号安全,请及时绑定邮箱和手机立即绑定

玩转数据结构之字典树(前缀树 Trie)

标签:
Java 算法

图片描述

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

图片描述

/// 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

图片描述

点击查看更多内容
1人点赞

若觉得本文不错,就分享一下吧!

评论

作者其他优质文章

正在加载中
PHP开发工程师
手记
粉丝
1.6万
获赞与收藏
1807

关注作者,订阅最新文章

阅读免费教程

感谢您的支持,我会继续努力的~
扫码打赏,你说多少就多少
赞赏金额会直接到老师账户
支付方式
打开微信扫一扫,即可进行扫码打赏哦
今天注册有机会得

100积分直接送

付费专栏免费学

大额优惠券免费领

立即参与 放弃机会
意见反馈 帮助中心 APP下载
官方微信

举报

0/150
提交
取消