【刷题-LeetCode】211. Add and Search Word - Data structure design

211. 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.

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.

解法1 hashmap。将单词按照长度映射

class WordDictionary {
public:
    unordered_map<int, set<string>>mp;
    /** Initialize your data structure here. */
    WordDictionary() {  }
    /** Adds a word into the data structure. */
    void addWord(string word) {
        int m = word.size();
        mp[m].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) {
        int m = word.size();
        if(mp.count(m) == 0)return false;
        for(auto s : mp[m]){
            int i;
            for(i = 0; i < m; ++i){
                if(s[i] == word[i] || word[i] == '.')continue;
                else break;
            }
            if(i == m)return true;
        }
        return false;
    }
};

解法2 trie-tree。查找时，对于包含了.的部分，递归往后查询

class trie_node{
public:
    bool isWord;
    vector<trie_node*>child;
    trie_node() : isWord(false), child(26, NULL){}
    ~trie_node(){
        for(auto &c : child)delete c;
    }
};
class WordDictionary {
public:
    trie_node* root;
    /** Initialize your data structure here. */
    WordDictionary() {
        root = new trie_node;
    }
    /** Adds a word into the data structure. */
    void addWord(string s) {
        trie_node *cur = root;
        for(int i = 0; i < s.size(); ++i){
            int idx = s[i] - 'a';
            if(cur->child[idx] == NULL){
                cur->child[idx] = new trie_node();
            }
            cur = cur->child[idx];
        }
        cur->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 _search(word, root);
    }
    bool _search(string s, trie_node* root){
        if(s == "")return root->isWord;
        if(s[0] == '.'){
            for(int j = 0; j < 26; ++j){
                if(root->child[j] && _search(s.substr(1), root->child[j]))return true;
            }
            return false;
        }else{
            if(root->child[s[0]-'a'] == NULL)return false;
            return _search(s.substr(1), root->child[s[0]-'a']);
        }
    }
};