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") -> falsesearch("bad") -> truesearch(".ad") -> truesearch("b..") -> true Note:
You may assume that all words are consist of lowercase lettersa-z. You should be familiar with how a Trie works. If not, please work on this problem: 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"); 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。