211. Add and Search Word - 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 .. A . means it can represent any one letter.

For 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.

Trie 总结: http://www.jianshu.com/p/357c1dbe31ec

Solution：Trie

思路： Trie基本套路 + "."跳过
1a: Iterative serach解法
1b: recursive search解法
Time Complexity: O(word_num * word_len)
Space Complexity: O(number of nodes)

Solution1a Code：

public class WordDictionary {
    public class TrieNode {
        public TrieNode[] children = new TrieNode[26];
        public boolean isWord; 
    }
    
    private TrieNode root = new TrieNode();

    public void addWord(String word) {
        TrieNode cur = root;
        for(int i = 0; i < word.length(); i++){
            char c = word.charAt(i);
            if(cur.children[c - 'a'] == null){
                cur.children[c - 'a'] = new TrieNode();
            }
            cur = cur.children[c - 'a'];
        }
        cur.isWord = true;
    }

    public boolean search(String word) {
        return search_helper(word, 0, root);
    }
    
    public boolean search_helper(String word, int start, TrieNode cur_node) {
        TrieNode cur = cur_node; 
        for(int i = start; i < word.length(); i++) {
            char c = word.charAt(i);
            if(c == '.') {
                for (int child = 0; child < cur.children.length; child++) {
                    if (cur.children[child] != null && search_helper(word, i + 1, cur.children[child])) {
                        return true;
                    }
                }
                return false;
            }
            else {
                if(cur.children[c - 'a'] == null) return false;
                cur = cur.children[c - 'a'];
            }
            
        }
        return cur.isWord;
    }
    
}

Solution1b Code：

public class WordDictionary {
    public class TrieNode {
        public TrieNode[] children = new TrieNode[26];
        public boolean isWord; 
    }
    
    private TrieNode root = new TrieNode();

    public void addWord(String word) {
        TrieNode cur = root;
        for(int i = 0; i < word.length(); i++){
            char c = word.charAt(i);
            if(cur.children[c - 'a'] == null){
                cur.children[c - 'a'] = new TrieNode();
            }
            cur = cur.children[c - 'a'];
        }
        cur.isWord = true;
    }

    public boolean search(String word) {
        return search_helper(word, 0, root);
    }

    
    public boolean search_helper(String word, int start, TrieNode cur_node) {
        TrieNode cur = cur_node; 
        for(int i = start; i < word.length(); i++) {
            char c = word.charAt(i);
            if(c == '.') {
                for (int child = 0; child < cur.children.length; child++) {
                    if (cur.children[child] != null && search_helper(word, i + 1, cur.children[child])) {
                        return true;
                    }
                }
                return false;
            }
            else {
                return cur.children[c - 'a'] != null && search_helper(word, start + 1, cur.children[c - 'a']);
            }
            
        }
        return cur.isWord;
    }
    
}