題意： 設計個字典查詢系統， 有 add 和search 兩種操作， add 是加入單詞到字典裡， search 時 可以用 點號萬用字元 "."， 點號可以匹配一個字母。

分析： 當search 時為 萬用字元時， 如果直接用back tracking產生 a-z, 比如 有7個點號， 就得生成  26^7 個組合，會TLE。 

以下是TLE的code: 

class WordDictionary {

    /** Initialize your data structure here. */
    Set<String> dic;
    
 
public WordDictionary() {
       dic = new HashSet<>();  
    }
    
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        dic.add(word);
    }
    
    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. 
 
*/
    public boolean search(String word) {
        
        return dfs(new StringBuilder(), word, 0);
    }
    
    private boolean dfs(StringBuilder curResult, String word, int index){
        if(curResult.length() == word.length()){
           // System.out.println(curResult.toString());
            if
 
(dic.contains(curResult.toString())) return true;
            return false;
        }
        
        boolean success = false;
        char cur_ch = word.charAt(index);    
        if(cur_ch == '.'){
            for(int i=0; i<26; i++){
                curResult.append((char)(i+'a'));
                success = success || dfs(curResult,word,index+1);
                curResult.setLength(curResult.length()-1);
                if(success) return true;
            }
        }
        
        else {
           curResult.append(cur_ch);
           success =  success || dfs(curResult,word,index+1);
           curResult.setLength(curResult.length()-1);
           if(success) return true; 
        }
        
        return success;
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

 改進: 用map 來存放 <長度+ List<String> > 的組合， 匹配一個單詞 首先得長度匹配，才能進一步匹配。

換成如下 演算法能beat 99%, 但如果單詞長度全部一樣， 那 變成了 n^2的演算法了。主要還是測試資料太弱了。

class WordDictionary {

    /** Initialize your data structure here. */
    Map<Integer, List<String>> dict;
    public WordDictionary() {
       dict = new HashMap<>();  
    }
    
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        List<String> val = dict.getOrDefault(word.length(),new ArrayList<>()) ;
        val.add(word);
        dict.put(word.length(), val);
    }
    
    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    public boolean search(String word) {
        if(!dict.containsKey(word.length())) return false; 
       
        List<String> list = dict.get(word.length());
        
        int i;
        for(String str: list){
           
            for(i=0; i<word.length(); i++){
                char c = word.charAt(i);
                if(c == '.') continue;
                if(c != str.charAt(i)) break;
            }
            if(i == word.length()) return true;
        }
        return false;
    }
}

Trie 的演算法 待續。