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.

这道题目考察的还是Trie Tree的问题，所以如果做了Implement Trie (Prefix Tree)【详情可以参考：http://blog.csdn.net/zuoyexingchennn/article/details/46315595】这道题目，这道题应该不难，唯一不同的地方就是search时候有些不同，这里，只要细心一些，应该就没问题。

AP程序如下：
【注】：这里的类名和leetcode原题的类名略有不同！

public class AddAndSearchWordDataStructureDesign {    private TrieNode root;    public AddAndSearchWordDataStructureDesign () {        root = new TrieNode();    }    public AddAndSearchWordDataStructureDesign(TrieNode root){        this.root=root;    }    // Adds a word into the data structure.    public void addWord(String word) {        TrieNode cur=root;        for(int i=0;i<word.length();i++){            int index=word.charAt(i)-'a';            if(cur.children[index]==null){                cur.children[index]=new TrieNode(word.charAt(i));            }            cur=cur.children[index];        }        cur.wordEnd=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) {        TrieNode cur=root;        for(int i=0;i<word.length();i++){            if(word.charAt(i)=='.'){                if(i==word.length()-1){                    for(int j=0;j<26;j++){                        if(cur.children[j]!=null&&cur.children[j].wordEnd)                            return true;                    }                    return false;                }                for(int j=0;j<26;j++){                    if(cur.children[j]!=null&&new AddAndSearchWordDataStructureDesign(cur.children[j]).search(word.substring(i+1)))                        return true;                }                return false;            }else{                int index=word.charAt(i)-'a';                if(cur.children[index]==null){                    return false;                }                cur=cur.children[index];                        }        }        if(!cur.wordEnd)            return false;        return true;    }    class TrieNode{           public TrieNode[] children=new TrieNode[26];;            public char c;            public boolean wordEnd;            public TrieNode() {            }            public TrieNode(char c){                this.c=c;            }    }}