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
解题思路：
参考之前的Java for LeetCode 208 Implement Trie (Prefix Tree) 修改下即可，JAVA实现如下：
public class WordDictionary extends Trie {
public void addWord(String word) {
super.insert(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) {
if (word == null || word.length() == 0)
return false;
return search(word, 0, root);
}
public boolean search(String word, int depth, TrieNode node) {
if (depth == word.length() - 1) {
if (word.charAt(depth) != '.') {
if (node.son[word.charAt(depth) - 'a'] != null) {
node = node.son[word.charAt(depth) - 'a'];
return node.isEnd;
} else
return false;
}
for (int i = 0; i < 26; i++) {
if (node.son[i] != null) {
TrieNode ason = node.son[i];
if (ason.isEnd)
return true;
}
}
return false;
}
if (word.charAt(depth) != '.') {
if (node.son[word.charAt(depth) - 'a'] != null) {
node = node.son[word.charAt(depth) - 'a'];
return search(word, depth + 1, node);
} else
return false;
}
for (int i = 0; i < 26; i++) {
if (node.son[i] != null) {
TrieNode ason = node.son[i];
if (search(word, depth + 1, ason))
return true;
}
}
return false;
}
}
class TrieNode {
// Initialize your data structure here.
int num;// 有多少单词通过这个节点,即节点字符出现的次数
TrieNode[] son;// 所有的儿子节点
boolean isEnd;// 是不是最后一个节点
char val;// 节点的值
TrieNode() {
this.num = 1;
this.son = new TrieNode[26];
this.isEnd = false;
}
}
class Trie {
protected TrieNode root;
public Trie() {
root = new TrieNode();
}
public void insert(String word) {
if (word == null || word.length() == 0)
return;
TrieNode node = this.root;
char[] letters = word.toCharArray();
for (int i = 0; i < word.length(); i++) {
int pos = letters[i] - 'a';
if (node.son[pos] == null) {
node.son[pos] = new TrieNode();
node.son[pos].val = letters[i];
} else {
node.son[pos].num++;
}
node = node.son[pos];
}
node.isEnd = true;
}
// Returns if the word is in the trie.
public boolean search(String word) {
if (word == null || word.length() == 0) {
return false;
}
TrieNode node = root;
char[] letters = word.toCharArray();
for (int i = 0; i < word.length(); i++) {
int pos = letters[i] - 'a';
if (node.son[pos] != null) {
node = node.son[pos];
} else {
return false;
}
}
return node.isEnd;
}
// Returns if there is any word in the trie
// that starts with the given prefix.
public boolean startsWith(String prefix) {
if (prefix == null || prefix.length() == 0) {
return false;
}
TrieNode node = root;
char[] letters = prefix.toCharArray();
for (int i = 0; i < prefix.length(); i++) {
int pos = letters[i] - 'a';
if (node.son[pos] != null) {
node = node.son[pos];
} else {
return false;
}
}
return true;
}
}
// Your Trie object will be instantiated and called as such:
// Trie trie = new Trie();
// trie.insert("somestring");
// trie.search("key");