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.

Solution:

public class WordDictionary {
public class TrieNode{
TrieNode[] childs;
boolean hasWord;
public TrieNode(){
childs = new TrieNode[26];
hasWord = false;
}
} TrieNode root; public WordDictionary(){
root = new TrieNode();
} // Adds a word into the data structure.
public void addWord(String word) {
TrieNode cur = root;
for (char c : word.toCharArray()){
if (cur.childs[c-'a']==null){
cur.childs[c-'a'] = new TrieNode();
}
cur = cur.childs[c-'a'];
}
cur.hasWord = true;
} // 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 searchRecur(root,word,0);
} public boolean searchRecur(TrieNode curNode, String word, int start){
if (curNode==null){
return false;
} if (start>=word.length()){
return curNode.hasWord;
} char c = word.charAt(start);
if (c=='.'){
for (TrieNode child : curNode.childs)
if (searchRecur(child,word,start+1)){
return true;
}
return false;
} else {
return searchRecur(curNode.childs[c-'a'],word,start+1);
}
}
} // Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");

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