Medium
Design a data structure that supports adding new words and finding if a string matches any previously added string.
Implement the WordDictionary class:
WordDictionary() Initializes the object.void addWord(word) Adds word to the data structure, it can be matched later.bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter.Example:
Input
["WordDictionary","addWord","addWord","addWord","search","search","search","search"] [[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
Output
[null,null,null,null,false,true,true,true]
Explanation
WordDictionary wordDictionary = new WordDictionary();
wordDictionary.addWord("bad");
wordDictionary.addWord("dad");
wordDictionary.addWord("mad");
wordDictionary.search("pad"); // return False
wordDictionary.search("bad"); // return True
wordDictionary.search(".ad"); // return True
wordDictionary.search("b.."); // return True
Constraints:
1 <= word.length <= 500word in addWord consists lower-case English letters.word in search consist of '.' or lower-case English letters.50000 calls will be made to addWord and search.public class WordDictionary {
private class Node {
public Node[] kids = new Node[26];
public bool isTerminal;
}
private readonly Node[] root = new Node[26];
public WordDictionary() {
// empty constructor
}
public void AddWord(string word) {
int n = word.Length;
if (root[n] == null) {
root[n] = new Node();
}
Node node = root[n];
for (int i = 0; i < n; i++) {
int c = word[i] - 'a';
Node kid = node.kids[c];
if (kid == null) {
kid = new Node();
node.kids[c] = kid;
}
node = kid;
}
node.isTerminal = true;
}
public bool Search(string word) {
Node node = root[word.Length];
return node != null && Dfs(0, node, word);
}
private bool Dfs(int i, Node node, string word) {
int len = word.Length;
if (i == len) {
return false;
}
char c = word[i];
if (c == '.') {
foreach (Node kid in node.kids) {
if (kid == null) {
continue;
}
if ((i == len - 1 && kid.isTerminal) || Dfs(i + 1, kid, word)) {
return true;
}
}
return false;
}
Node next = node.kids[c - 'a'];
if (next == null) {
return false;
}
if (i == len - 1) {
return next.isTerminal;
}
return Dfs(i + 1, next, word);
}
}
}
/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary obj = new WordDictionary();
* obj.AddWord(word);
*/