[Leetcode] 288. Unique Word Abbreviation 解题报告

题目：

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)     1b) d|o|g                   --> d1g              1    1  1     1---5----0----5--8c) i|nternationalizatio|n  --> i18n              1     1---5----0d) 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") -> falseisUnique("cart") -> trueisUnique("cane") -> falseisUnique("make") -> true

思路：

用哈希表来存储一个缩写对应的单词列表。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，然后去哈希表中查找。如果哈希表中这个缩写对应几个单词，那么肯定不唯一；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；如果这个缩写在哈希表中没有对应单词，那么肯定是唯一的了。感觉这个题目有点无聊。

代码：

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;          }        else if(hash[abb].size() == 1) {            return word == hash[abb][0];        }        else {            return true;        }     }private:    unordered_map<string, vector<string>> hash;};/** * Your ValidWordAbbr object will be instantiated and called as such: * ValidWordAbbr obj = new ValidWordAbbr(dictionary); * bool param_1 = obj.isUnique(word); */