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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 }
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原文地址：http://www.cnblogs.com/warmland/p/5962591.html
