288. Unique Word Abbreviation

Posted 2020-08-14 warmland

这是一题非常坑的简单题。

https://discuss.leetcode.com/topic/37254/let-me-explain-the-question-with-better-examples/2

1) [“dog”]; isUnique(“dig”);   

//False, because “dig” has the same abbreviation with “dog" and “dog” is already in the dictionary. It’s not unique.

2) [“dog”, “dog"]; isUnique(“dog”);  

//True, because “dog” is the only word that has “d1g” abbreviation.

3) [“dog”, “dig”]; isUnique(“dog”);   

//False, because if we have more than one word match to the same abbreviation, this abbreviation will never be unique.

这样就能明白了

 

　思路:

　　建一个Hashmap，key是缩写，value是对应的词，如果对词典进行建立的时候就已经发现重复了，就把value设置成“”

　　对新来的词，如果这个词的缩写在map并且这个词不是那个字典里产生这个词的原词，那么就返回false。

 1 public class ValidWordAbbr {
 2     Map<String, String> dic;
 3 
 4     public ValidWordAbbr(String[] dictionary) {
 5         dic = new HashMap<String, String>();
 6         for(int i = 0; i < dictionary.length; i++) {
 7             String abb = generateAbb(dictionary[i]);
 8             if(dic.containsKey(abb) && !dic.get(abb).equals(dictionary[i])) {
 9                 dic.put(abb, "");
10             } else {
11                 dic.put(abb, dictionary[i]);
12             }
13         }
14     }
15     
16     private String generateAbb(String s) {
17         int len = s.length();
18         if(len < 3) {
19             return s;
20         }
21         StringBuilder sb = new StringBuilder();
22         sb.append(s.charAt(0));
23         sb.append(len - 2);
24         sb.append(s.charAt(len - 1));
25         return sb.toString();
26     }
27 
28     public boolean isUnique(String word) {
29         String abb = generateAbb(word);
30         return !dic.containsKey(abb) || dic.get(abb).equals(word);
31     }
32 }