Add and Search Word

题目描述：

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

.
解题思路：使用字典树(trie)

AC代码如下（时间：85msclass TrieNode {
public:
// Initialize your data structure here.
TrieNode(){
m_isWord = false;
for (int i = 0; i < 26; ++i){
next[i] = NULL;
}
}
void setIsWord(bool isWord){
m_isWord = isWord;
}
bool getIsWord(void){
return m_isWord;
}
public:
bool m_isWord;
TrieNode* next[26];
};

class WordDictionary {
public:
WordDictionary(){
root = new TrieNode();
}
// Adds a word into the data structure.
void addWord(string word) {
int len = word.length();
if (len == 0) return;
TrieNode* p = root;
for (int i = 0; i < len; ++i){
if (p->next[word[i] - 'a'] == NULL){
p->next[word[i] - 'a'] = new TrieNode();
p = p->next[word[i] - 'a'];
}
else{
p = p->next[word[i] - 'a'];
}
}
p->setIsWord(true);
}
bool search(string word){
return search(word, 0, root);
}
// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
bool search(const string& word,int curIndex,TrieNode* curNode) {
if (curIndex == word.length()) return curNode->getIsWord();
if (word[curIndex] == '.'){
bool res = false;
for (int i = 0; i < 26; ++i){
if (curNode->next[i] != NULL){
res = search(word, curIndex + 1, curNode->next[i]);
if (res) return true;
}
}
return false;
}
else{
if (curNode->next[word[curIndex] - 'a'] != NULL){
return search(word, curIndex + 1, curNode->next[word[curIndex] - 'a']);
}
else{
return false;
}
}
}
private:
TrieNode* root;
};

）