[Leetcode] 211. Add and Search Word - Data structure design 解题报告

题目：

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

.
思路：

这道题目的正解就是实现Trie这种数据结构（请参考我的博文《[Leetcode] 208. Implement Trie (Prefix Tree) 解题报告》）。唯一不同的地方在于本题目中出现了通配符‘.’，也就是说'.'可以通配a
4000
-z中的任意一个字符。因此在search中需要采用DFS进行穷举：一旦发现了通配符，则尝试将其替换成每一个可能的字符，一旦匹配成功就返回true；如果所有的匹配都失败，则返回false。

代码：

class TrieNode {
public:
bool isComplete;
TrieNode* ch[26];
TrieNode() {
isComplete = false;
for(int i = 0; i < 26; ++i) {
ch[i] = NULL;
}
}
};

class Trie {
public:
Trie() {
root = new TrieNode();
}

void insert(string word) {
TrieNode* p = root;
for(auto c : word) {
if(p->ch[c - 'a'] == NULL) {
p->ch[c - 'a'] = new TrieNode();
}
p = p->ch[c - 'a'];
}
p->isComplete = true;
}

bool search(string word) {
return dfs(root, word, 0);
}

bool dfs(TrieNode* p, string& word, int startIndex) {
if(p == NULL) {
return false;
}
if(startIndex == word.length()) {
return p->isComplete;
}
char c = word[startIndex];
if(c == '.') {
for(int i = 0; i < 26; ++i) {
if(p->ch[i] != NULL && dfs(p->ch[i], word, startIndex + 1)) {
return true;
}
}
return false;
}
else {
return dfs(p->ch[c-'a'], word, startIndex + 1);
}
}
private:
TrieNode* root;
};

class WordDictionary {
public:
/** Initialize your data structure here. */
WordDictionary() {

}

/** Adds a word into the data structure. */
void addWord(string word) {
trie.insert(word);
}

/** 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 trie.search(word);
}
private:
Trie trie;
};

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