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 --> l10nAssume 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
public class ValidWordAbbr {
Map<String, String> map = new HashMap<>();
public ValidWordAbbr(String[] dictionary) {
for (String dic : dictionary) {
String key = getKey(dic);
if (map.containsKey(key)) {
map.put(key, "");
} else {
map.put(key, dic);
}
}
}
public boolean isUnique(String word) {
String key = getKey(word);
return !map.containsKey(key) || map.get(key).equals(word);
}
private String getKey(String word) {
String key = word.charAt(0) + Integer.toString(word.length() - 2) + word.charAt(word.length() - 1);
return key;
}
}
// 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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