[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)

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") -> [code]false

isUnique("cart") ->

true

isUnique("cane") ->

false

isUnique("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);
*/