从Leetcode上学字典树

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面试 › 数据结构与算法

字典树（前缀树）Trie是一种数据结构，用于高效的存储和检索字符串数据集中的键。应用场景：自动补充和拼写检查

Leetcode.208 实现前缀树

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class Trie {

    private Trie[] children;
    private boolean isEnd;

    public Trie() {
        children = new Trie[26];
        isEnd = false;
    }
    
    public void insert(String word) {
        Trie node = this;
        for(int i=0;i<word.length();i++) {
            char ch = word.charAt(i);
            int index = ch - 'a';
            if(node.children[index]==null) {
                node.children[index] = new Trie();
            }
            node = node.children[index];
        }
        node.isEnd = true;
    }
    
    public boolean search(String word) {
        Trie node = searchPrefix(word);
        return node != null && node.isEnd;
    }
    
    public boolean startsWith(String prefix) {
        return searchPrefix(prefix) != null;
    }

    private Trie searchPrefix(String prefix) {
        Trie node = this;
        for(int i=0;i<prefix.length();i++) {
            char ch = prefix.charAt(i);
            int index = ch - 'a';
            if(node.children[index]==null) {
                return null;
            }
            node = node.children[index];
        }
        return node;
    }
}

/**
 * Your Trie object will be instantiated and called as such:
 * Trie obj = new Trie();
 * obj.insert(word);
 * boolean param_2 = obj.search(word);
 * boolean param_3 = obj.startsWith(prefix);
 */

这里使用了Trie数组来存储子节点，也可以使用TreeMap来实现

Leetcode.720 字典中最长的单词

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class Solution {
    public String longestWord(String[] words) {
        Trie trie = new Trie();
        for (String word : words) {
            trie.insert(word);
        }
        String longest = "";
        for (String word : words) {
            if (trie.search(word)) {
                if (word.length() > longest.length() || (word.length() == longest.length() && word.compareTo(longest) < 0)) {
                    longest = word;
                }
            }
        }
        return longest;
    }
}

class Trie {
    Trie[] children;
    boolean isEnd;

    public Trie() {
        children = new Trie[26];
        isEnd = false;
    }
    
    public void insert(String word) {
        Trie node = this;
        for (int i = 0; i < word.length(); i++) {
            char ch = word.charAt(i);
            int index = ch - 'a';
            if (node.children[index] == null) {
                node.children[index] = new Trie();
            }
            node = node.children[index];
        }
        node.isEnd = true;
    }
    
    public boolean search(String word) {
        Trie node = this;
        for (int i = 0; i < word.length(); i++) {
            char ch = word.charAt(i);
            int index = ch - 'a';
            if (node.children[index] == null || !node.children[index].isEnd) {
                return false;
            }
            node = node.children[index];
        }
        return node != null && node.isEnd;
    }
}

这题是对Trim的完美应用。精辟的在于search()函数中，一旦该字符串的节点路径上存在节点的isEnd属性为假，那么就将该字符串抛弃