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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.
click to show hint.
You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.
LeetCode出新题的速度越来越快了,有点跟不上节奏的感觉了。这道题如果做过之前的那道 Implement Trie (Prefix Tree) 实现字典树(前缀树)的话就没有太大的难度了,还是要用到字典树的结构,唯一不同的地方就是search的函数需要重新写一下,因为这道题里面'.'可以代替任意字符,所以一旦有了'.',就需要查找所有的子树,只要有一个返回true,整个search函数就返回true,典型的DFS的问题,其他部分跟上一道实现字典树没有太大区别,代码如下:
class WordDictionary {
public:
struct TrieNode {
public:
TrieNode *child[26];
bool isWord;
TrieNode() : isWord(false) {
for (auto &a : child) a = NULL;
}
};
WordDictionary() {
root = new TrieNode();
}
// Adds a word into the data structure.
void addWord(string word) {
TrieNode *p = root;
for (auto &a : word) {
int i = a - 'a';
if (!p->child[i]) p->child[i] = new TrieNode();
p = p->child[i];
}
p->isWord = true;
}
// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
bool search(string word) {
return searchWord(word, root, 0);
}
bool searchWord(string &word, TrieNode *p, int i) {
if (i == word.size()) return p->isWord;
if (word[i] == '.') {
for (auto &a : p->child) {
if (a && searchWord(word, a, i + 1)) return true;
}
return false;
} else {
return p->child[word[i] - 'a'] && searchWord(word, p->child[word[i] - 'a'], i + 1);
}
}
private:
TrieNode *root;
};
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");
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