Add and Search Word
Design a data structure that supports the following two operations: addWord(word) and 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:
addWord("bad")
addWord("dad")
addWord("mad")
search("pad") // return false
search("bad") // return true
search(".ad") // return true
search("b..") // return true
Solution:
We can create a trie to store all the data. During search, when we meet a dot, we will check all possible candidates on that level of the trie.
Code:
public class WordDictionary {
class TrieNode {
TrieNode[] next;
boolean isWord;
public TrieNode() {
next = new TrieNode[26];
isWord = false;
}
}
TrieNode root = new TrieNode();
public void addWord(String word) {
if (word == null || word.length() == 0) {
return;
}
TrieNode p = root;
for (int i = 0; i < word.length(); i++) {
int c = word.charAt(i) - 'a';
if (p.next[c] == null) {
p.next[c] = new TrieNode();
}
p = p.next[c];
}
p.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;
}
return find(word, 0, root);
}
private boolean find(String word, int i, TrieNode p) {
if (i == word.length()) {
return p.isWord;
}
char c = word.charAt(i);
if (c == '.') {
for (int j = 0; j < p.next.length; j++) {
if (p.next[j] != null && find(word, i + 1, p.next[j])) {
return true;
}
}
} else if (p.next[c - 'a'] != null) {
return find(word, i + 1, p.next[c - 'a']);
}
return false;
}
}