[LeetCode] 208. Implement Trie (prefix tree)

topic

A Trie (pronounced like "try") or  prefix tree  is a tree-like data structure used to efficiently store and retrieve keys from a string dataset. This data structure has quite a few applications, such as autocompletion and spell checking.

Please implement the Trie class:

• Trie() Initialize the prefix tree object.
• void insert(String word) Insert a string into the prefix tree  word .
• boolean search(String word) If the string  word is in the prefix tree, return it  true(that is, it was inserted before retrieval); otherwise, return it  false .
• boolean startsWith(String prefix) Returns  if word one of the prefixes of   the previously inserted string  is  ; otherwise, returns   .prefixtruefalse

Example:

enter
["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
[[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
output
[null, null, true, false, true, null, true]

explain
Trie trie = new Trie();
trie.insert("apple");
trie.search("apple"); // returns True
trie.search("app"); // returns False
trie.startsWith("app"); // returns True
trie.insert("app");
trie.search("app"); // returns True

hint:

• 1 <= word.length, prefix.length <= 2000
• word and  prefix consist only of lowercase English letters
• insert, search and   the totalstartsWith  number of calls   does not exceed 3 * 10^4

answer

source code

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) {
        return searchPrefix(word) != null && searchPrefix(word).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);
 */

Summarize

It's a bit complicated... the first time I didn't bother to give up, and the second time I got AC.

The official solution is actually equivalent to a 26-fork number. The member variable children array represents the next letter, the 0 index corresponds to 'a', and the 25 index corresponds to 'z'; isEnd indicates whether the current letter is the end of a word.