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211. Add and Search Word Data structure design
Jacky Zhang edited this page Dec 5, 2016
·
1 revision
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.
解题思路为Trie。
public class WordDictionary {
public class TrieNode {
public TrieNode[] children = new TrieNode[26];
public String item = "";
}
private TrieNode root = new TrieNode();
// Adds a word into the data structure.
public void addWord(String word) {
TrieNode node = root;
for (char c : word.toCharArray()) {
if (node.children[c - 'a'] == null) {
node.children[c - 'a'] = new TrieNode();
}
node = node.children[c - 'a'];
}
node.item = word;
}
// 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 match(word.toCharArray(), 0, root);
}
private boolean match(char[] chs, int k, TrieNode node) {
if(k == chs.length) return !node.item.equals("");
if (chs[k] != '.') {
return node.children[chs[k] - 'a'] != null && match(chs, k+1, node.children[chs[k] - 'a']);
} else {
for(int i = 0; i < node.children.length; i++) {
if(node.children[i] != null) {
if(match(chs, k+1, node.children[i])) return true;
}
}
}
return false;
}
}
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");