[leetcode] 288. Unique Word Abbreviation 解题报告

题目链接: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

思路: 将字典中的单词按照缩写为key, 将具有相同缩写的放进一个hash表中. 当判断一个给定单词是否唯一的时候,可以先计算出其缩写, 然后去hash表中查, 如果hash表中这个缩写对应几个单词, 那么肯定是不唯一的, 如果只对应一个单词, 则看这个单词是不是我们要判断的单词, 如果是的话,则唯一, 否则不唯一. 如果这个缩写在hash表中没有对应单词, 那么肯定唯一了.

代码如下:

class ValidWordAbbr {
public:
    ValidWordAbbr(vector<string> &dictionary) {
        for(auto str: dictionary)
        {
            if(str.size() <= 2) continue;
            int num = str.size()-2;
            string abb = str[0] + to_string(num) + str[str.size()-1];
            hash[abb].push_back(str);
        }
    }

    bool isUnique(string word) {
        int num = word.size()-2;
        string abb = word[0] + to_string(num) + word[word.size()-1];
        if(hash[abb].size() > 1) return false;
        if(hash[abb].size() == 1) return word==hash[abb][0];
        return true;
    }
private:
    unordered_map<string, vector<string>> hash;
};


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