Add and Search Word with Regular Expression - Data structure design

Add and Search Word with Regular Expression - Data structure design

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 . or * . A . means it can represent any one letter. A * can be any sequence of characters (including the empty sequence).

For example:

addWord("bad")

addWord("dad")

addWord("mad")

search("pad") -> false

search("bad") -> true

search(".ad") -> true

search("b..") -> true

Thoughts

empty意味着i+1但node不往后走, any other sequence意味着node可以跳过任意次, 再和i + 1匹配.

Code

class WordDictionary {
    class TrieNode {
        TrieNode[] children = new TrieNode[26];
        boolean end;
    }
    TrieNode root;
    /** Initialize your data structure here. */
    public WordDictionary() {
        root = new TrieNode();
    }

    /** Adds a word into the data structure. */
    public void addWord(String word) {
        TrieNode node = root;
        for (int i = 0; i < word.length(); i++) {
            int index = word.charAt(i) - 'a';
            if (node.children[index] == null) {
                node.children[index] = new TrieNode();
            }
            node = node.children[index];
        }
        node.end = true;
    }

    private boolean dfs(String word, TrieNode node, int pos) {
        for (int i = pos; i < word.length(); i++) {
            if (word.charAt(i) == '.') {
                for (int j = 0; j < 26; j++) {
                    if (node.children[j] != null) {
                        if (dfs(word, node.children[j], i + 1)) {
                            return true;
                        }
                    }
                }
                return false;
            } else if (word.charAt(i) == '*') {
                // be any seq of characters, including empty
                for (int j = 0; j < 26; j++) {
                    if (node.children[j] != null) {
                        if (dfs(word, node.children[j], i) // skip nodes: acccb(node), a*(pos)b skip ccc
                            || dfs(word, node.children[j], i + 1) // jump to next character, a*b(pos), acccb(node)
                            || dfs(word, node, i + 1)) { // skip *: empty a(node)b, a*(pos)b => a(node)b, a*b(pos)
                            return true;
                        } 
                    }
                }
                return false;
            } else {
                int index = word.charAt(i) - 'a';
                if (node.children[index] == null) {
                    return false;
                }
                node = node.children[index];
            }
        }
        return node.end;
    }

    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    public boolean search(String word) {
        return dfs(word, root, 0);
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

Analysis

时间复杂度指数级.