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.
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.
这题是Trie的基本操作。
我的经验是把Trie的操作提取出来,像这里面的 getOrCreate这个函数。
这样逻辑就简化很多了。
其它的就挺直接的这里就不多说了。
class WordDictionary {
TrieNode root;
/** Initialize your data structure here. */
public WordDictionary() {
root = new TrieNode('a');
}
/** Adds a word into the data structure. */
public void addWord(String word) {
TrieNode cur = root;
for (char c : word.toCharArray()) {
cur = cur.getOrCreate(c);
}
cur.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) {
return helper(root, word, 0);
}
private boolean helper(TrieNode root, String word, int index) {
if (root == null) return false;
if (index == word.length()) {
return root.isWord;
}
if (word.charAt(index) != '.') {
return helper(root.get(word.charAt(index)), word, index + 1);
} else {
for (char c = 'a'; c <= 'z'; c++) {
if (helper(root.get((char) c), word, index + 1)) return true;
}
return false;
}
}
}
class TrieNode {
char c;
boolean isWord;
TrieNode[] child;
public TrieNode(char c) {
this.c = c;
child = new TrieNode[26];
}
public TrieNode getOrCreate(char c) {
if (child[c - 'a'] == null) {
child[c - 'a'] = new TrieNode(c);
}
return get(c);
}
public TrieNode get(char c) {
return child[c - 'a'];
}
}
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