*LeetCode-Add and Search Word - Data structure design

这个题的trie tree解法以后要记得看

首先我想的每次add一个词 就add bad/ .ad / ..d / b.d太慢了

应该是用map存 用长度做key 一个list存这个长度的所有string

然后每次首先判断长度 然后判断是否是完整的词 List有contains函数！！！

然后假如不是词 而是有.的 就一位一位判断

public class WordDictionary {
    HashMap<Integer,List<String>> dict = new HashMap<Integer, List<String>>();
    // Adds a word into the data structure.
    public void addWord(String word) {
        if ( dict.containsKey( word.length() )){
            List<String> list = dict.get( word.length() );
            if ( !list.contains( word ) )
                list.add ( word );
        }
        else{
            List<String> temp = new ArrayList<String>();
            temp.add ( word );
            dict.put ( word.length(), temp );
        }
    }
    // 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() );
        if ( isWord ( word ) ){
            return list.contains( word );
        }
        else {
            for ( String str: list ){
                if ( same ( word, str ) )
                    return true;
            }
        }
        return false;
    }
    public boolean isWord ( String word ){
        for ( int i = 0; i < word.length(); i++ ){
            if ( !Character.isLetter( word.charAt(i) ) )
                return false;
        }
        return true;
    }
    public boolean same ( String word, String str){
        for ( int i = 0; i < word.length(); i ++ ) {
            if ( word.charAt(i) != str.charAt(i) && word.charAt(i) != '.')
                return false;
        }
        return true;
    }
}