LeetCode #288. Unique Word Abbreviation

最新推荐文章于 2021-02-10 07:12:36 发布

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本文介绍了一种算法，用于判断给定单词在特定词典中的缩写是否唯一。通过创建一个包含词典中所有单词缩写的映射，可以快速检查任意单词的缩写是否与词典中其他单词的缩写重复。此方法适用于需要高效判断单词缩写唯一性的场景。

题目描述：

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

class ValidWordAbbr {
public:
    ValidWordAbbr(vector<string>& dictionary) {
        for(string& s:dictionary) dict[abbreviation(s)].insert(s);
    }
    
    bool isUnique(string word) {
        string s=abbreviation(word);
        // 词典中要么没有这个缩写，要么只有一个相同的单词
        return dict[s].count(word)==dict[s].size();
    }
    
    string abbreviation(string& s)
    {
        if(s.size()>2) return s.front()+to_string(s.size()-2)+s.back();
        else return s;
    }
    
private:
    unordered_map<string,unordered_set<string>> dict;   
};