需求

■如何判断一堆不重复的字符串是否以某个前缀开头?
用Set\Map存储字符串
遍历所有字符串进行判断
时间复杂度：O(n)

■有没有更优的数据结构实现前缀搜索?

Trie

◼Trie 也叫做字典树、前缀树(Prefix Tree)、单词查找树
◼Trie 搜索字符串的效率主要跟字符串的长度有关
◼假设使用 Trie 存储 cat、dog、doggy、does、cast、add 六个单词

接口设计

接口一：

int size();
boolean isEmpty();
void clear();
boolean contains(String str);
void add(String str);
void remove(String str);
boolean starsWith(String prefix);

接口二：

int size();
boolean isEmpty();
void clear();
boolean contains(String str);
V add(String str, V value);
V remove(String str);
boolean starsWith(String prefix);

实现

Trie类

package alangeit;

import java.util.HashMap;

public class Trie<V> {
    private int size;
    private Node<V> root;
    
    public int size() {
        return size;
    }

    public boolean isEmpty() {
        return size == 0;
    }

    public void clear() {
        size = 0;
        root = null;
    }

    public V get(String key) {
        Node<V> node = node(key);
        return node != null && node.word ? node.value : null;
    }

    public boolean contains(String key) {
        Node<V> node = node(key);
        return node != null && node.word;
    }

    public V add(String key, V value) {
        keyCheck(key);
        
        // 创建根节点
        if (root == null) {
            root = new Node<>(null);
        }

        Node<V> node = root;
        int len = key.length();
        for (int i = 0; i < len; i++) {
            char c = key.charAt(i); 
            boolean emptyChildren = node.children == null;
            Node<V> childNode = emptyChildren ? null : node.children.get(c);
            if (childNode == null) {
                childNode = new Node<>(node);
                childNode.character = c;
                node.children = emptyChildren ? new HashMap<>() : node.children;
                node.children.put(c, childNode);
            }
            node = childNode;
        }
        
        if (node.word) { // 已经存在这个单词
            V oldValue = node.value;
            node.value = value;
            return oldValue;
        }
        
        // 新增一个单词
        node.word = true;
        node.value = value;
        size++;
        return null;
    }

    public V remove(String key) {
        // 找到最后一个节点
        Node<V> node = node(key);
        // 如果不是单词结尾，不用作任何处理
        if (node == null || !node.word) return null;
        size--;
        V oldValue = node.value;
        
        // 如果还有子节点
        if (node.children != null && !node.children.isEmpty()) {
            node.word = false;
            node.value = null;
            return oldValue;
        }
        
        // 如果没有子节点
        Node<V> parent = null;
        while ((parent = node.parent) != null) {
            parent.children.remove(node.character);
            if (parent.word || !parent.children.isEmpty()) break;
            node = parent;
        }
        
        return oldValue;
    }

    public boolean startsWith(String prefix) {
        return node(prefix) != null;
    }
    
    private Node<V> node(String key) {
        keyCheck(key);
        
        Node<V> node = root;
        int len = key.length();
        for (int i = 0; i < len; i++) {
            if (node == null || node.children == null || node.children.isEmpty()) return null;
            char c = key.charAt(i); 
            node = node.children.get(c);
        }
        
        return node;
    }
    
    private void keyCheck(String key) {
        if (key == null || key.length() == 0) {
            throw new IllegalArgumentException("key must not be empty");
        }
    }
    
    private static class Node<V> {
        Node<V> parent;
        HashMap<Character, Node<V>> children;
        Character character;
        V value;
        boolean word; // 是否为单词的结尾（是否为一个完整的单词）
        public Node(Node<V> parent) {
            this.parent = parent;
        }
    }
}

Asserts类

package alangeit;

public class Asserts {
    public static void test(boolean value) {
        try {
            if (!value) throw new Exception("测试未通过");
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

测试

static void test() {
    Trie<Integer> trie = new Trie<>();
    trie.add("cat", 1);
    trie.add("dog", 2);
    trie.add("catalog", 3);
    trie.add("cast", 4);
    trie.add("小码哥", 5);
    Asserts.test(trie.size() == 5);
    Asserts.test(trie.startsWith("do"));
    Asserts.test(trie.startsWith("c"));
    Asserts.test(trie.startsWith("ca"));
    Asserts.test(trie.startsWith("cat"));
    Asserts.test(trie.startsWith("cata"));
    Asserts.test(!trie.startsWith("hehe"));
    Asserts.test(trie.get("小码哥") == 5);
    Asserts.test(trie.remove("cat") == 1);
    Asserts.test(trie.remove("catalog") == 3);
    Asserts.test(trie.remove("cast") == 4);
    Asserts.test(trie.size() == 2);
    Asserts.test(trie.startsWith("小"));
    Asserts.test(trie.startsWith("do"));
    Asserts.test(!trie.startsWith("c"));
}

总结

◼Trie 的优点:搜索前缀的效率主要跟前缀的长度有关

◼Trie 的缺点:需要耗费大量的内存，因此还有待改进

◼更多Trie 相关的数据结构和算法
Double-array Trie、Suffix Tree、Patricia Tree、Crit-bit Tree、AC自动机