[LeetCode] Add and Search Word - Data structure design

Add and Search Word - Data structure design

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 letters a-z.

解题思路：

运用前缀树来实现词典的查询。唯一不同的是有通配符.，对于通配符，运用暴力回溯搜索法即可。

struct dictionaryNode{public:     dictionaryNode* sons[26];    bool isWord;    dictionaryNode(){        for(int i=0; i<26; i++){            sons[i]=NULL;        }        isWord=false;    }    ~dictionaryNode(){        for(int i=0; i<26; i++){            if(sons[i]!=NULL){                delete sons[i];                sons[i]=NULL;            }        }    }};class WordDictionary {public:    dictionaryNode* root;        WordDictionary(){        root=new dictionaryNode();    }        ~WordDictionary(){        if(root!=NULL){            delete root;            root = NULL;        }        }    // Adds a word into the data structure.    void addWord(string word) {        dictionaryNode* node = root;        int len=word.length();        for(int i=0; i<len; i++){            if(node->sons[word[i]-'a']==NULL){                node->sons[word[i]-'a'] = new dictionaryNode();            }            node = node->sons[word[i]-'a'];        }        node->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 searchHelp(root, word, 0);    }        bool searchHelp(dictionaryNode* node, string& word, int i){        if(node==NULL){            return false;        }        int len = word.length();        if(i>=len){            return node->isWord;        }                if(word[i]!='.'){            return searchHelp(node->sons[word[i]-'a'], word, i+1);        }else{            for(int k=0; k<26; k++){                if(searchHelp(node->sons[k], word, i+1)){                    return true;                }            }            return false;        }    }};// Your WordDictionary object will be instantiated and called as such:// WordDictionary wordDictionary;// wordDictionary.addWord("word");// wordDictionary.search("pattern");