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 解题关键点有3个:

1. 找出word abbreviation 的规律,<first letter><number><last letter>,number = string.length() - 2

2. 当发现dictionary 里有相同的abbreviation, key 对应的value 变为""

3. The abbreviation of "hello", i.e., h3o already exists in the dictionary.

Input: ["hello"],isUnique("hello") Output: [false] Expected: [true]

If the given word itself is in the dictionary, and it has the unique abbreviation, then we should return true.

 public class ValidWordAbbr {
private Map<String, String> map = new HashMap<String, String>(); public ValidWordAbbr(String[] dictionary) {
for(int i = ; i < dictionary.length; i++){
String key = abbreviate(dictionary[i]);
if(!map.containsKey(key)){
map.put(key, dictionary[i]);
}else{
map.put(key, "");
}
}
} private String abbreviate(String str){
return str.charAt() + Integer.toString(str.length() - )+ str.charAt(str.length()-);
} public boolean isUnique(String word) {
String x = abbreviate(word);
if(map.containsKey(x)){
if(map.get(x).equals(word)){
return true;
}else {
return false;
}
}
return true;
}
} // Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");

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