[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") -> false
search("bad") -> true
search(".ad") -> true
search("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");


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