An abbreviation of a word follows the form <first letter><number><last letter>. Below are some examples of word abbreviations:

a) it                      --> it    (no abbreviation)

     1
b) d|o|g                   --> d1g

              1    1  1
     1---5----0----5--8
c) i|nternationalizatio|n  --> i18n

              1
     1---5----0
d) l|ocalizatio|n          --> l10n

Assume you have a dictionary and given a word, find whether its abbreviation is unique in the dictionary. A word's abbreviation is unique if no other word from the dictionary has the same abbreviation.
Example:

Given dictionary = [ "deer", "door", "cake", "card" ]
isUnique("dear") -> 
false
isUnique("cart") -> 
true
isUnique("cane") -> 
false
isUnique("make") -> 
true

Solution：hashmap

思路：前提遍历存好(abr_key -> str)的map，如果有重复，将str置为""使之无效这样要判断的同key的str和它一定不相等，return false。
Time Complexity: 初始化构造O(N) isUnique(): O(1) Space Complexity: O(N)

Solution Code：

public class ValidWordAbbr {
    HashMap<String, String> map;
    public ValidWordAbbr(String[] dictionary) {
        map = new HashMap<String, String>();
        for(String str: dictionary){
            String key = getKey(str);
            // If there is more than one string belong to the same key
            // then the key will be invalid, we set the value to ""
            if(map.containsKey(key)){
                if(!map.get(key).equals(str)){
                    map.put(key, "");
                }
            }
            else{
                map.put(key, str);
            }
        }
    }

    public boolean isUnique(String word) {
        return !map.containsKey(getKey(word)) || map.get(getKey(word)).equals(word);
    }
    
    String getKey(String str){
        if(str.length() <= 2) return str;
        return str.charAt(0)+Integer.toString(str.length() - 2)+str.charAt(str.length() - 1);
    }
}