题目：
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
链接： http://leetcode.com/problems/unique-word-abbreviation/
题解：
新题的题目真是越来越长了。 这道题是给定一个数组Dictionary， 求输入字符串是否有unique的abbreviation在Dictionary中。我们选择用Map<String, Set<String>>来做我们存储数据的数据结构，然后按照题意做就可以了，还需要判断一些边界条件，比如缩写不在map中直接返回true之类的。 以后一定要牢记，选定了好的数据结构，编写程序就会容易很多。
Time Complexity - O(n * L)， Space Complexity - O(n * L)。
public class ValidWordAbbr {
private Map<String, HashSet<String>> map;
public ValidWordAbbr(String[] dictionary) {
this.map = new HashMap<>();
for(int i = 0; i < dictionary.length; i++) {
String abbr = getAbbr(dictionary[i]);
if(!map.containsKey(abbr)) {
HashSet<String> set = new HashSet<>();
set.add(dictionary[i]);
map.put(abbr, set);
} else {
if(!map.get(abbr).contains(dictionary[i])) {
map.get(abbr).add(dictionary[i]);
}
}
}
}
public boolean isUnique(String word) {
if(map.size() == 0 || word.length() < 3) {
return true;
}
String abbr = getAbbr(word);
if(!map.containsKey(abbr) || (map.get(abbr).contains(word) && map.get(abbr).size() == 1)) {
return true;
} else {
return false;
}
}
private String getAbbr(String s) {
if(s.length() < 3) {
return s;
} else {
return s.substring(0, 1) + String.valueOf(s.length() - 2) + s.substring(s.length() - 1);
}
}
}
// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");
二刷:
跟一刷的方法一样。就是跟Anagram一样，用Map<String, Set<String>>来存，使用一个新的方法getAbbr先求出abbr作为key，然后把单词加入到key的value里。 Discuss里面还有很多很好的方法，用map<String, String>之类的，三刷要好好研究。
Java:
Time Complexity - O(n * L)， Space Complexity - O(n * L)。
public class ValidWordAbbr {
Map<String, Set<String>> map;
public ValidWordAbbr(String[] dictionary) {
map = new HashMap<>();
for (String s : dictionary) {
String abbr = getAbbr(s);
if (!map.containsKey(abbr)) {
map.put(abbr, new HashSet<String>());
}
map.get(abbr).add(s);
}
}
public boolean isUnique(String word) {
String abbr = getAbbr(word);
if (!map.containsKey(abbr) || (map.get(abbr).contains(word) && map.get(abbr).size() == 1)) {
return true;
}
return false;
}
private String getAbbr(String s) {
if (s.length() < 3) {
return s;
}
int len = s.length();
return s.substring(0, 1) + (len - 2) + s.substring(len - 1);
}
}
// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");
Reference:
https://leetcode.com/discuss/62842/a-simple-java-solution-using-map-string-string
https://leetcode.com/discuss/61658/share-my-java-solution
https://leetcode.com/discuss/71652/java-solution-with-hashmap-string-string-beats-submissions