[Locked] Unique Word Abbreviation

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

分析：

　　其实题目没有表达清楚...应该包含如下意思：

　　1. dictionary = {"dear"},  isUnique("door") -> false

　　2. dictionary = {"door", "door"}, isUnique("door") -> true

　　3. dictionary = {"dear", "door"}, isUnique("door") -> false

　　所以当缩写存在时，也并非一定要return false，如果原字典中与query缩写一致的原字符串，如2中dict的两个"door"，与query原字符串"door"一致，那么也应该return true。

代码：

class Solution {
private:
    string maptoabbr(string str) {
        string abbr = "";
        abbr += str[0];
        //若只有一位，则直接返回；有两位，则中间不加数字；两个以上，则加数字
        if(str.length() > 1) {
            abbr += str.length() > 2 ? to_string(str.length() - 2) : "";
            abbr += str.back();
        }
        return abbr;
    }
    //hashabbr用来存储缩写后的字符串，hashorig用来存储原始字符串
    unordered_multiset<string> hashabbr, hashorig;
    
public:
    Solution(vector<string> dict) {
        for(string str : dict) {
            hashorig.insert(str);
            hashabbr.insert(maptoabbr(str));
        }
    }
    bool isUnique(string str) {
        string abbr = maptoabbr(str);
        //如果缩写不存在字典中，直接return true
        if(hashabbr.find(abbr) == hashabbr.end())
            return true;
        //如果缩写在字典中，则如果query只对应一种原始字符串，则return true；否则return false
        return hashabbr.count(abbr) == hashorig.count(str);
    }
};