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
Show Company Tags
Show Tags
Show Similar Problems
 
 class ValidWordAbbr {
private:
unordered_map<string, int> dict;
unordered_set<string> inDict;
string getAbbr(string word) {
if (word.size() < ) return word;
return word.front() + std::to_string(word.size() - ) + word.back();
}
public:
ValidWordAbbr(vector<string> &dictionary) {
for (int i = , n = dictionary.size(); i < n; i++) {
if (inDict.count(dictionary[i])) continue;
inDict.insert(dictionary[i]);
string abbr = getAbbr(dictionary[i]);
if (dict.count(abbr) == )
dict.insert(make_pair(abbr, ));
else dict[abbr]++;
}
} bool isUnique(string word) {
string abbr = getAbbr(word);
if (dict.count(abbr) == || (dict[abbr] == && inDict.count(word) == ))
return true;
return false;
}
}; // Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa(dictionary);
// vwa.isUnique("hello");
// vwa.isUnique("anotherWord");

要点:int转string可以用函数std::to_string()

inDict 的作用有两个:1)将所给的字典去重。 2)判断要检测的单词是否在所给的字典中。

Unique Word Abbreviation -- LeetCode的更多相关文章

  1. [LeetCode] Minimum Unique Word Abbreviation 最短的独一无二的单词缩写

    A string such as "word" contains the following abbreviations: ["word", "1or ...

  2. [Locked] Unique Word Abbreviation

    Unique Word Abbreviation An abbreviation of a word follows the form <first letter><number&g ...

  3. Leetcode Unique Word Abbreviation

    An abbreviation of a word follows the form <first letter><number><last letter>. Be ...

  4. 288. Unique Word Abbreviation

    题目: An abbreviation of a word follows the form <first letter><number><last letter> ...

  5. Unique Word Abbreviation

    An abbreviation of a word follows the form <first letter><number><last letter>. Be ...

  6. [LeetCode] Unique Word Abbreviation 独特的单词缩写

    An abbreviation of a word follows the form <first letter><number><last letter>. Be ...

  7. [LeetCode] 288.Unique Word Abbreviation 独特的单词缩写

    An abbreviation of a word follows the form <first letter><number><last letter>. Be ...

  8. Leetcode: Minimum Unique Word Abbreviation

    A string such as "word" contains the following abbreviations: ["word", "1or ...

  9. [Swift]LeetCode288. 唯一单词缩写 $ Unique Word Abbreviation

    An abbreviation of a word follows the form <first letter><number><last letter>. Be ...

随机推荐

  1. windows下使用RoboCopy命令进行文件夹增量备份

    RoboCopy,它是一个命令行的目录复制命令,自从Windows NT 4.0 开始就成为windows 资源工具包的一部分,然后在Windows Vista.Windows 7和 Windows ...

  2. Eclipse 出现“polling news feeds”的解决办法

    小编突然心血来潮,安装了一下Java的环境,eclipse的IDE来写点Java,但是是不是出现以下的弹窗,实在是闹心,后来网上看了前辈们的解决办法,特此记录一下.如有侵权,敬请告知!!! 1. 找到 ...

  3. CSU-2214 Sequence Magic

    题目链接 http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2214 题目 Description 有一个1到N的自然数序列1,2,3, ...

  4. ERC720和erc721的区别

    有一阵子,Ethereum网络突然变的特别拥堵,原因是兴起了一款以太坊养猫的Dapp游戏,超级可爱的猫形象,再加上配种,繁殖和拍卖等丰富的玩法,风靡了币圈. 一时间币圈大大小小的人都在撸猫,以太坊网络 ...

  5. Struts2+DAO层实现实例03——添加监听器跟踪用户行为

    实例说明 根据上两次的成品进行二次加工. 加入Listener,监听用户的登陆注销情况. 所用知识说明 采用SessionBindingListener对Session进行监听. 同时,Action中 ...

  6. Struts2+DAO层实现实例01——搭建Struts2基本框架

    实例内容 利用Strust2实现一个登陆+注册功能的登陆系统. 实现基础流程:

  7. Java UDP的简单实例以及知识点简述

    UDP的实现 Java中实现UDP协议的两个类,分别是DatagramPacket数据包类以及DatagramSocket套接字类. 其与TCP协议实现不同的是: UDP的套接字DatagramSoc ...

  8. Android记事本开发01

    今天: 学习一下Android的基本知识,了解一下记事本开发大概需要哪些知识. 昨天: 无 遇到的问题:

  9. P3531 [POI2012]LIT-Letters

    题目描述 Little Johnny has a very long surname. Yet he is not the only such person in his milieu. As it ...

  10. [洛谷P4149][IOI2011]Race

    题目大意:给一棵树,每条边有边权.求一条简单路径,权值和等于$K$,且边的数量最小. 题解:点分治,考虑到这是最小值,不满足可减性,于是点分中的更新答案的地方计算重复的部分要做更改,就用一个数组记录前 ...