211. Add and Search Word - Data structure design

Posted 2021-02-19 wentiliangkaihua

Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)

search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

Example:

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

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

public class WordDictionary {
    class TrieNode {
        TrieNode[] child = new TrieNode[26];
        boolean isWord = false;
    }
    TrieNode root = new TrieNode();
    public void addWord(String word) {
        TrieNode p = root;
        for (char c : word.toCharArray()) {
            if (p.child[c - ‘a‘] == null) p.child[c - ‘a‘] = new TrieNode();
            p = p.child[c - ‘a‘];
        }
        p.isWord = true;
    }

    public boolean search(String word) {
        return doSearch(root, word.toCharArray(), 0);
    }
    
    private boolean doSearch(TrieNode cur, char[] word, int pos) {
        if (cur == null) return false;
        if (pos == word.length) return cur.isWord;
        
        if (word[pos] == ‘.‘) {
            for (TrieNode next : cur.child)
                if (doSearch(next, word, pos + 1))
                    return true;
            return false;
        }
        return doSearch(cur.child[word[pos] - ‘a‘], word, pos + 1);
    }
}

Trie Trie Trie

 private boolean helper(String s, int index, TrieNode p) {
        if (index >= s.length()) return p.isWord;
        char c = s.charAt(index);
        if (c == ‘.‘) {
            for (int i = 0; i < p.child.length; i++)
                if (p.child[i] != null && helper(s, index + 1, p.child[i]))
                    return true;
            return false;
        } else return (p.child[c - ‘a‘] != null && helper(s, index + 1, p.child[c - ‘a‘]));
    }

search的method也可以这样