Add and Search Word
Design a data structure that supports the following two operations:
void addWord(word)
boolean 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..") -> trueNote: You may assume that all words are consist of lowercase letters a-z.
Analysis
在Implement Trie的基础之上，应用Trie。不同之处在于对于pattern的匹配，需要进行backtracking，也就是利用DFS，对于匹配了'.'的那个TrieNode的每一个children都循环遍历，如果有一个为true，则说明找到一个match，即可返回true。
Solution
class TrieNode {
public boolean isLeaf;
public TrieNode[] children;
public TrieNode() {
this.children = new TrieNode[26];
}
}
public class WordDictionary {
private TrieNode root;
public WordDictionary() {
this.root = new TrieNode();
}
// Adds a word into the data structure.
public void addWord(String word) {
TrieNode p = this.root;
for (int i = 0; i < word.length(); i++) {
int index = word.charAt(i) - 'a';
if (p.children[index] == null) {
TrieNode temp = new TrieNode();
p.children[index] = temp;
}
p = p.children[index];
}
p.isLeaf = 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) {
return match(this.root, word, 0);
}
private boolean match(TrieNode p, String word, int k) {
if (k == word.length()) {
return p.isLeaf;
}
if (word.charAt(k) == '.') {
for (int i = 0; i < p.children.length; i++) {
if (p.children[i] != null) {
if (match(p.children[i], word, k + 1)) {
return true;
}
}
}
} else {
int index = word.charAt(k) - 'a';
return p.children[index] != null && match(p.children[index], word, k + 1);
}
return false;
}
}
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");(126ms 54.49%)
class WordDictionary {
public class TrieNode {
private TrieNode[] children;
private boolean isEnd;
public TrieNode() {
children = new TrieNode[26];
}
public TrieNode[] getChildren() {
return children;
}
public boolean containsKey(char ch) {
return children[ch - 'a'] != null;
}
public TrieNode get(char ch) {
return children[ch - 'a'];
}
public void put(char ch, TrieNode node) {
children[ch - 'a'] = node;
}
public void setEnd() {
isEnd = true;
}
public boolean isEnd() {
return isEnd;
}
}
private TrieNode root;
/** Initialize your data structure here. */
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 ch = word.charAt(i);
if (!node.containsKey(ch)) {
node.put(ch, new TrieNode());
}
node = node.get(ch);
}
node.setEnd();
}
/** 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, 0, root);
}
private boolean match(String word, int index, TrieNode node) {
if (index == word.length()) return node.isEnd();
char currentChar = word.charAt(index);
if (currentChar == '.') {
TrieNode[] children = node.getChildren();
for (int i = 0; i < children.length; i++) {
if (children[i] != null && match(word, index + 1, children[i])) {
return true;
};
}
} else {
return node.get(currentChar) != null && match(word, index + 1, node.get(currentChar));
}
return false;
}
}
/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary obj = new WordDictionary();
* obj.addWord(word);
* boolean param_2 = obj.search(word);
*/Reference
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