LeetCode 288. Unique Word Abbreviation（单词缩写）

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本文介绍了一种算法，用于判断一个单词的缩写形式在给定的词典中是否唯一。通过两种方法实现，一种是使用哈希映射存储每个缩写的单词，另一种是优化后的哈希映射，使用自定义的Abbr类来记录单词出现的次数。

原题网址：https://leetcode.com/problems/unique-word-abbreviation/

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

方法：比较直接的方法，就是建立哈希映射，以缩写为key，以原始的单词为value。

public class ValidWordAbbr {
    private Map<String, Set<String>> abbrs = new HashMap<>();

    private String abbr(String word) {
        if (word.length() <= 2) return word;
        StringBuilder sb = new StringBuilder();
        sb.append(word.charAt(0));
        sb.append(word.length()-2);
        sb.append(word.charAt(word.length()-1));
        return sb.toString();
    }
    public ValidWordAbbr(String[] dictionary) {
        for(String word: dictionary) {
            String abbr = abbr(word);
            Set<String> words = abbrs.get(abbr);
            if (words == null) {
                words = new HashSet<>();
                abbrs.put(abbr, words);
            }
            words.add(word);
        }
    }

    public boolean isUnique(String word) {
        Set<String> words = abbrs.get(abbr(word));
        if (words == null || words.isEmpty()) return true;
        if (words.size() == 1 && words.contains(word)) return true;
        return false;
    }
}

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

方法二：可以优化不使用哈希集合。

public class ValidWordAbbr {
    private Map<String, Abbr> map = new HashMap<>();
    
    private String abbr(String word) {
        if (word == null || word.length() <=2) return word;
        return word.substring(0,1) + Integer.toString(word.length()-2) + word.substring(word.length()-1);
    }
    
    public ValidWordAbbr(String[] dictionary) {
        for(String word: dictionary) {
            String key = abbr(word);
            Abbr abbr = map.get(key);
            if (abbr == null) abbr = new Abbr(word); 
            else if (!word.equals(abbr.word)) abbr.count ++;
            map.put(key, abbr);
        }
    }

    public boolean isUnique(String word) {
        Abbr abbr = map.get(abbr(word));
        if (abbr == null) return true;
        if (abbr.count > 1) return false;
        if (abbr.word.equals(word)) return true;
        return false;
    }
}
class Abbr {
    int count;
    String word;
    Abbr(String word) {
        this.word = word;
        this.count = 1;
    }
}

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