LeetCode211. Add and Search Word - Data structure design前缀树(Trie,字典树)实战

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

.
struct TrieNode
{
bool hasValue = false;
TrieNode *children[26];
TrieNode() {
memset(children, 0, sizeof(children));
}
~ TrieNode() {

}
};

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

void delTrie(TrieNode* node) {
for (int i = 0; i < 26; i++) {
if (node->children[i]) {
delTrie(node->children[i]);
}
}
delete node;
}

void addWord(const string& word)
{
TrieNode* node = root;
for (int i = 0; i < word.size(); i++) {
auto ch = node->children[word[i] - 'a'];
if (ch == nullptr) {
node->children[word[i] - 'a'] = new TrieNode();
}
node = node->children[word[i] - 'a'];
}
node->hasValue = true;
}

bool search(const string& word) {
return search(root, word);
}

private:
bool search(TrieNode* rootNode, const string& word) {
TrieNode* node = rootNode;
if (node == nullptr) {
return false;
}
for (int i = 0; i < word.size(); i++) {
if (word[i] == '.') {
for (int j = 0; j < 26; j++) {
if (node->children[j]) {
if (i == word.size() - 1)
{
if (node->children[j]->hasValue) {
return true;
}
else {
continue;
}
}
else if (search(node->children[j], word.substr(i + 1))) {
return true;
}
}
}
return false;
}
else {
if (node->children[word[i] - 'a'] == nullptr) {
return false;
}
else {
node = node->children[word[i] - 'a'];
}
}
}
return node->hasValue;
}
TrieNode *root;
};