211. Add and Search Word - Data structure design

Design a data structure that supports the following two operations: addWord(word) andsearch(word)

search(word) can search a literal word or a regular expression string containing only lettersa-z or ..

A . means it can represent any one letter.

 Notice

You may assume that all words are consist of lowercase letters a-z.

Have you met this question in a real interview? 

Yes

Example

addWord("bad")
addWord("dad")
addWord("mad")
search("pad")  // return false
search("bad")  // return true
search(".ad")  // return true
search("b..")  // return true

使用Trie

public class WordDictionary {
    private TrieNode root = new TrieNode(' ');
    public void addWord(String word) {
        TrieNode current = root;
        for (int i = 0; i < word.length(); i++) {
            char ch = word.charAt(i);
            TrieNode node = current.getChildNode(ch);
            if (node == null) {
                current.children.put(ch, new TrieNode(ch));
                node = current.getChildNode(ch);
            }
            current = node;
        }
        current.isEnd = true;
    }

    // Returns if the word is in the trie.
    public boolean search(String word) {
        return helper(root, word, 0);
    }

    public boolean helper(TrieNode current, String word, int index) {
        if (current == null) return false;
        if (index == word.length()) return current.isEnd;
        
        char c = word.charAt(index);
        if (c == '.') {
            for (TrieNode child : current.children.values()) {
                if (helper(child, word, index + 1)) return true;
            }
        } else {
            return helper(current.getChildNode(word.charAt(index)), word, index + 1);
        }
        return false;
    }
}

class TrieNode {
    char ch;
    boolean isEnd;
    Map<Character, TrieNode> children;

    public TrieNode(char ch) {
        this.ch = ch;
        children = new HashMap<>();
    }

    public TrieNode getChildNode(char ch) {
        return children.get(ch);
    }
}