211. Add and Search Word - Data structure design

摘要： Design a data structure that supports the following two operations: void addWord(word) bool search(word) search(word) can search a l...

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.

難度：medium

題目：

設計資料結構支援以下兩操作：

void addWord(word)

bool search(word)

search(word) 可以查單詞或包含.的正則表示式。 .可以表示任何單個字元。

注意：

你可以假定所有單詞僅由小寫字母組成。

思路：

Tire Tree, DFS

Runtime: 94 ms, faster than 89.57% of Java online submissions for Add and Search Word - Data structure design.

class WordDictionary {
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) {
char c = word.charAt(0);
int i = 0, idx = 0;
TrieNode ptr = root;
for (; i < word.length(); i++, ptr = ptr.next[idx]) {
idx = word.charAt(i) - 'a';
if (null == ptr.next[idx]) {
ptr.next[idx] = new TrieNode();
}
}

ptr.isWord = true;
}

// dfs 
/** 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 search(word, 0, root);
}

public boolean search(String word, int idx, TrieNode beginNode) {
if (idx >= word.length() || null == beginNode) {
return false;
}

TrieNode node = null;
int c = word.charAt(idx), cIdx = c - 'a';
if ('.' == c) {
boolean result = false;
for (int i = 0; i < 26; i++) {
node = beginNode.next[i];
if (null == node) {
continue;
}
if ((idx + 1) == word.length() && node.isWord) {
return true;
}
result = result || search(word, idx + 1, beginNode.next[i]);
}
return result;
}

// c is not equal to '.'
node = beginNode.next[cIdx];
if (null == node) {
return false;
}
if ((idx + 1) == word.length()) {
return node.isWord;
}

return search(word, idx + 1, beginNode.next[cIdx]);
}

public static class TrieNode {
public boolean isWord;
public TrieNode[] next = new TrieNode[26];
}
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */