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
     
       dic; 3  4     public ValidWordAbbr(String[] dictionary) { 5         dic = new HashMap
      
       (); 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 }