211. Add and Search Word - Data structure design

211. Add and Search Word - Data structure design

题目：

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.

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.

代码如下：

class TrieNode {
public:
bool isKey;
TrieNode* children[26];
TrieNode(): isKey(false) {
memset(children, NULL, sizeof(TrieNode*) * 26);
}
};

class WordDictionary {
public:
WordDictionary() {
root = new TrieNode();
}

// Adds a word into the data structure.
void addWord(string word) {
TrieNode* run = root;
for (char c : word) {
if (!(run -> children[c - 'a']))
run -> children[c - 'a'] = new TrieNode();
run = run -> children[c - 'a'];
}
run -> isKey = true;
}

// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
bool search(string word) {
return query(word.c_str(), root);
}

private:
TrieNode* root;

bool query(const char* word, TrieNode* node) {
TrieNode* run = node;
for (int i = 0; word[i]; i++) {
if (run && word[i] != '.')
run = run -> children[word[i] - 'a'];
else if (run && word[i] == '.') {
TrieNode* tmp = run;
for (int j = 0; j < 26; j++) {
run = tmp -> children[j];
if (query(word + i + 1, run))
return true;
}
}
else break;
}
return run && run -> isKey;
}
};