[leetcode] 288. Unique Word Abbreviation 解题报告
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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) 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") ->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");
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