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  • 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
    

    Analyse: no other word from the dictionary has the same abbreviation means that if the dictionary includes that word, and the abbreviation of that word exist only once, then we should return true. hash_map + set

    Analyse: hash_map, map the abbreviated word to its original word. If the the abbreviated word already exist in the hash map, and the value of it does not equals to the word itself, set it as empty. Otherwise, map the abbreviated and the original word. 

    Runtime: 177ms

     1 class ValidWordAbbr {
     2 public:
     3     ValidWordAbbr(vector<string> &dictionary) {
     4         for (string word : dictionary) {
     5             string abbreviatedWord = getAbbreviate(word);
     6             // more than two words map to a single abbreviate word
     7             if (abbreviations[abbreviatedWord] != "" && abbreviations[abbreviatedWord] != word) 
     8                 abbreviations[abbreviatedWord] = "";
     9             else
    10                 abbreviations[abbreviatedWord] = word;
    11         }
    12     }
    13 
    14     bool isUnique(string word) {
    15          string abbreviatedWord = getAbbreviate(word);
    16          return abbreviations.count(abbreviatedWord) == 0 || abbreviations[abbreviatedWord] == word;
    17     }
    18     
    19 private:
    20     unordered_map<string, string> abbreviations;
    21     
    22     string getAbbreviate(string word) {
    23         int n = word.size();
    24         if (n < 3) return word;
    25         else  return word[0] + to_string(n - 2) + word[n - 1];
    26     }
    27 };
    28 
    29 
    30 // Your ValidWordAbbr object will be instantiated and called as such:
    31 // ValidWordAbbr vwa(dictionary);
    32 // vwa.isUnique("hello");
    33 // vwa.isUnique("anotherWord");

    Analyse: hash_map + set

    Runtime: 255ms

     1 class ValidWordAbbr {
     2 public:
     3     ValidWordAbbr(vector<string> &dictionary) {
     4         for (string word : dictionary)
     5             abbreviateAndOriginal[getAbbreviate(word)].insert(word);
     6     }
     7 
     8     bool isUnique(string word) {
     9         unordered_set<string> mapToMe = abbreviateAndOriginal[getAbbreviate(word)];
    10         if (mapToMe.empty()) return true;
    11         return mapToMe.size() == 1 && mapToMe.count(word) == 1;
    12     }
    13 private:
    14     unordered_map<string, unordered_set<string> > abbreviateAndOriginal;
    15     
    16     string getAbbreviate(string word) {
    17         int n = word.size();
    18         if (n < 3) return word;
    19         else 
    20             return word[0] + to_string(n - 2) + word[n - 1];
    21     }
    22 };

    Analyse: two hash_map

    Runtime: 233ms 

     1 class ValidWordAbbr {
     2 public:
     3     ValidWordAbbr(vector<string> &dictionary) {
     4         for (string word : dictionary) {
     5             if (!wordExist[word])
     6                 abbreviations[getAbbreviation(word)]++;
     7             wordExist[word] = true;
     8         }
     9     }
    10 
    11     bool isUnique(string word) {
    12         return (wordExist[word] && abbreviations[getAbbreviation(word)] == 1) ||
    13                (!wordExist[word] && !abbreviations[getAbbreviation(word)]);
    14     }
    15 private:
    16     unordered_map<string, bool> wordExist;
    17     unordered_map<string, int> abbreviations;
    18     
    19     string getAbbreviation(string word) {
    20         int n = word.size();
    21         if (n < 3) return word;
    22         else return word[0] + to_string(n - 2) + word[n - 1];
    23     }
    24 };

    Analyse: hash_map + vector

    Runtime: 295ms

     1 class ValidWordAbbr {
     2 public:
     3     ValidWordAbbr(vector<string> &dictionary) {
     4         for (string word : dictionary) {
     5             string findMe = getAbbreviate(word);
     6             if (find(abbreviations[findMe].begin(), abbreviations[findMe].end(), word) == abbreviations[findMe].end())
     7                 abbreviations[findMe].push_back(word);
     8         }
     9     }
    10 
    11     bool isUnique(string word) {
    12         vector<string> words = abbreviations[getAbbreviate(word)];
    13         return words.empty() ||
    14                (words.size() == 1 && words[0] == word);
    15                
    16     }
    17     
    18 private:
    19     unordered_map<string, vector<string> > abbreviations;
    20     
    21     string getAbbreviate(string word) {
    22         int n = word.size();
    23         if (n < 3) return word;
    24         else  return word[0] + to_string(n - 2) + word[n - 1];
    25     }
    26 };
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/5894792.html
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