*LeetCode-Unique Word Abbreviation

这个题也很简单 但是要想全可能存在的情况 所以假如只有hello 那hello是可以的 或者根本dict里面没有这个词 也没有这个词的abb也是可以的

我用了map of string and int 来count有几个这样的abb 还用了一个set来记录dict里面的词 这样很浪费

简便方法是用abb做key 但是value就用dict里面的词本身 假如第二次碰到这个abb了 那么把value设置成“” 空 最后就判断哪个word和value是否相等就知道是否只有一个了 不相等就是并不是只有一个

public class ValidWordAbbr {
    HashMap<String, Integer> map = new HashMap<String, Integer>();
    Set<String> words = new HashSet<String>();
    public ValidWordAbbr(String[] dictionary) {
        for ( int i = 0; i < dictionary.length; i ++ ){
            String abb = "";
            int len = dictionary[ i ].length();
            if ( len <= 2 )
                abb = dictionary[ i ];
            else
                abb = dictionary[ i ].charAt( 0 ) + Integer.toString(len - 2) + dictionary[ i ].charAt( len - 1 );
            if ( map.containsKey( abb ) )
                map.put ( abb, 2 );
            else
                map.put ( abb, 1 );
            words.add ( dictionary[ i ] );
        }
    }

    public boolean isUnique(String word) {
        String abb = "";
        int len = word.length();
        if ( len <= 2 )
            abb = word;
        else
            abb = word.charAt( 0 ) + Integer.toString(len - 2) + word.charAt( len - 1 );
        if ( !map.containsKey ( abb ) )
            return true;
        else if ( words.contains ( word ) && map.get( abb ) == 1 )
            return true;
        return false;
    }
}

public class ValidWordAbbr {
    HashMap<String, String> map = new HashMap<String, String>();
    public ValidWordAbbr(String[] dictionary) {
        for ( int i = 0; i < dictionary.length; i ++ ){
            String abb = "";
            int len = dictionary[ i ].length();
            if ( len <= 2 )
                abb = dictionary[ i ];
            else
                abb = dictionary[ i ].charAt( 0 ) + Integer.toString(len - 2) + dictionary[ i ].charAt( len - 1 );
            if ( map.containsKey( abb ) )
                map.put ( abb, "" );
            else
                map.put ( abb, dictionary[ i ] );
        }
    }

    public boolean isUnique(String word) {
        String abb = "";
        int len = word.length();
        if ( len <= 2 )
            abb = word;
        else
            abb = word.charAt( 0 ) + Integer.toString(len - 2) + word.charAt( len - 1 );
        if ( !map.containsKey ( abb ) )
            return true;
        if ( map.get( abb ).equals ( word ))
            return true;
        return false;
    }
}