Add and Search Word - Data structure design
Desicription
Design a data structure that supports the following two operations:
代码语言：javascript复制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.
Example:
代码语言：javascript复制addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> trueNote:
You may assume that all words are consist of lowercase letters a-z.
Solution
代码语言：javascript复制/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary obj = new WordDictionary();
* obj.addWord(word);
* bool param_2 = obj.search(word);
*/
class WordDictionary {
private:
class TrieNode {
public:
bool isWord = false;
TrieNode* children[26] = {nullptr};
};
TrieNode* root = new TrieNode();
public:
/** Initialize your data structure here. */
WordDictionary() = default;
/** Adds a word into the data structure. */
void addWord(string word) {
TrieNode* run = root;
for(char c : word) {
if(nullptr == run->children[c - 'a'])
run->children[c - 'a'] = new TrieNode();
run = run->children[c - 'a'];
}
run->isWord = 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(std::move(word), root);
}
bool query(string word, TrieNode* run, int index = 0) {
for(int i = index; word[i]; i ) {
if(nullptr != run && word[i] != '.')
run = run->children[word[i] - 'a'];
else if(nullptr != run && word[i] == '.') {
TrieNode* tmp = run;
for(char c = 'a'; c <= 'z'; c ) {
word[i] = c;
run = tmp->children[word[i] - 'a'];
if(query(word, run, i 1))
return true;
}
}
else break;
}
return run && run->isWord;
}
};
