Question Definition
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.
You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.
Java Solution
class WordDictionary {
private TrieNode root = new TrieNode();
private class TrieNode {
private final int R = 26; // radix = 26
public TrieNode[] next;
public boolean isWord;
public TrieNode() {
next = new TrieNode[R];
}
}
// Adds a word into the data structure.
public void addWord(String word) {
if(word == null || word.length() == 0)
return;
TrieNode node = root;
int d = 0;
while(d < word.length()) {
char c = word.charAt(d);
if(node.next[c - 'a'] == null)
node.next[c - 'a'] = new TrieNode();
node = node.next[c - 'a'];
d++;
}
node.isWord = true;
}
// 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) {
if(word == null || word.length() == 0)
return false;
TrieNode node = root;
int d = 0;
return search(node, word, 0);
}
private boolean search(TrieNode node, String word, int d) {
if(node == null)
return false;
if(d == word.length())
return node.isWord;
char c = word.charAt(d);
if(c == '.') {
for(TrieNode child : node.next) {
if(child != null && search(child, word, d + 1))
return true;
}
return false;
} else {
return search(node.next[c - 'a'], word, d + 1);
}
}
}
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