Unique Word Abbreviation

这是一道设计题，请好好理解题意。一开始，我们设计的是只用Set，提交了才知道，hello提交给hello是返回true，遂改成Map来解。

public class ValidWordAbbr {

Map<String, Set<String>> set;

public ValidWordAbbr(String[] dictionary) {
set = new HashMap<>();
for (String str: dictionary) {
String abbString = getAbbString(str);
if (set.containsKey(abbString)) {
Set<String> subSet = set.get(abbString);
subSet.add(str);
} else {
Set<String> subSet = new HashSet<>();
subSet.add(str);
set.put(abbString, subSet);
}
}
}

public boolean isUnique(String word) {
String abbString = getAbbString(word);
if (set.containsKey(abbString)) {
Set<String> subSet = set.get(abbString);
if (subSet.contains(word) && subSet.size() == 1) {
return true;
} else {
return false;
}
} else {
return true;
}
}

private String getAbbString(String word) {
if (word.length() <= 2) {
return word;
} else {
int length = word.length() - 2;
String abbStr = word.substring(0, 1) + String.valueOf(length) + word.substring(word.length() - 1);
return abbStr;
}
}
}

// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");