Add and Search Word
211. Add and Search Word - Data structure design
LintCode-473.Add and Search Word
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.
class TrieNode {
private var children: [TrieNode?] = []
public var isEnd = false
init() {
self.children = Array(repeating: nil, count: 26)
self.isEnd = false
}
func put(_ c: Character, _ node: TrieNode) {
let val = Int(c.unicodeScalars.first!.value - UnicodeScalar("a").value)
children[val] = node
}
func get(_ c: Character) -> TrieNode? {
return children[Int(c.unicodeScalars.first!.value - UnicodeScalar("a").value)]
}
func getChildren() -> [TrieNode?] {
return children
}
}
class WordDictionary {
private var root: TrieNode
/** Initialize your data structure here. */
init() {
self.root = TrieNode()
}
/** Adds a word into the data structure. */
func addWord(_ word: String) {
var node = root
for c in word {
if let n = node.get(c) {
node = n
} else {
let n = TrieNode()
node.put(c, n)
node = n
}
}
node.isEnd = true
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
func search(_ word: String) -> Bool {
return search(word, root, 0)
}
func search(_ word: String, _ node: TrieNode?, _ i: Int) -> Bool {
guard let node = node else {
return false
}
if i == word.count {
return node.isEnd
}
let c = word[word.index(word.startIndex, offsetBy: i)]
if c == "." {
for n in node.getChildren() where n != nil {
if search(word, n, i + 1) {
return true
}
}
return false
} else {
return search(word, node.get(c), i + 1)
}
}
}
/**
* Your WordDictionary object will be instantiated and called as such:
* let obj = WordDictionary()
* obj.addWord(word)
* let ret_2: Bool = obj.search(word)
*/
let dict = WordDictionary()
dict.addWord("bad")
dict.addWord("dad")
dict.addWord("mad")
print(dict.search("pad"))
print(dict.search("bad"))
print(dict.search(".ad"))
print(dict.search("b.."))
class TrieNode {
private TrieNode[] children;
private boolean isEnd;
public TrieNode() {
children = new TrieNode[26];
}
public boolean containsKey(char c) {
return children[c -'a'] != null;
}
public TrieNode get(char c) {
return children[c -'a'];
}
public void put(char c, TrieNode node) {
children[c -'a'] = node;
}
public void setEnd() {
isEnd = true;
}
public boolean isEnd() {
return isEnd;
}
public TrieNode[] children() {
return children;
}
}
public class WordDictionary {
private TrieNode root;
public WordDictionary() {
root = new TrieNode();
}
// Adds a word into the data structure.
public void addWord(String word) {
TrieNode node = root;
for (int i = 0; i < word.length(); i++) {
char c = word.charAt(i);
if (!node.containsKey(c)) {
node.put(c, new TrieNode());
}
node = node.get(c);
}
node.setEnd();
}
private boolean search(TrieNode node, String word, int index) {
if (node == null) {
return false;
} else if (index == word.length()) {
return node.isEnd();
}
char c = word.charAt(index);
if (c == '.') {
for (TrieNode n : node.children()) {
if (n != null) {
if(search (n, word, index + 1)) {
return true;
}
}
}
return false;
} else {
TrieNode n = node.get(c);
return search(n, word, index + 1);
}
}
// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
public boolean search(String word) {
return search(root, word, 0);
}
}
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");
Hope this helps,
Michael