leetcode.211. Add and Search Word - Data structure design字典树

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 lettersa-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 letters a-z.

用到字典树

 class TrieNode     {  public:      bool isEnd;      TrieNode *children[26];      TrieNode() : isEnd(false)      {          for (int i = 0; i < 26; i++)          {              children[i] = NULL;          }      }  }; class WordDictionary {public:    WordDictionary()      {          root = new TrieNode();      }            // Adds a word into the data structure.    void addWord(string word) {        TrieNode *cur = root;          for (int i = 0; i < word.length(); i++)          {              int index = word[i] - 'a';              if (cur->children[index] == NULL)              {                  cur->children[index] = new TrieNode();              }                cur = cur->children[index];          }            cur->isEnd = 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) {        int n = word.length();          return search(word, n, 0, root);     }    bool search(string &word, int n, int pos, TrieNode *cur)      {          if (cur == NULL)          {              return false;          }          if (pos == n)          {              return cur->isEnd;          }            if (word[pos] == '.')          {              for (int i = 0; i < 26; i++)              {                  if (cur->children[i])                  {                      if (search(word, n, pos+1, cur->children[i]))                      {                          return true;                      }                  }              }          }          else          {              int index = word[pos] - 'a';              if (cur->children[index])              {                  return search(word, n, pos+1, cur->children[index]);              }          }            return false;      }  private:      TrieNode *root;  };