Xiaohe-LeetCode 288 Unique Word Abbreviation-优快云博客

本文深入探讨了一个用于判断单词缩写是否唯一的算法实现。通过实例分析和错误案例，作者展示了如何优化该算法以提高效率，并指出使用字符串长度而非大小写的潜在效率提升。文章详细介绍了算法的核心思想，包括如何利用映射表来存储和查找缩写，以及如何正确处理不同长度的输入单词以确保唯一性。此外，还讨论了几个具体的测试用例，以验证算法的正确性和效率。

This question is confusing. But if we see the wrong cases below, we can understand the meaning.

If word's abbr is not in dic, then return true.

Else if word's abbr is in dic

　　{The matching words in dic are all exactly the same word to the input word: return true. else return false}

Else return false;

(1)Right Solution:(need to find better solution)

59.64%,256ms. The interesting parts is that if I use d.size() instead of d.length() here, the efficiency will drop down to

292 ms, 29.2%.

class ValidWordAbbr {
public:
ValidWordAbbr(vector<string> &dictionary) {
for (string& d : dictionary) {
int n = d.length();
string abbr = d[0] + to_string(n) + d[n - 1];
mp[abbr].insert(d);
}
}

bool isUnique(string word) {
int n = word.length();
string abbr = word[0] + to_string(n) + word[n - 1];
return mp[abbr].count(word) == mp[abbr].size();
}
private:
unordered_map<string, unordered_set<string>> mp;
};

(2) Wrong Case:

Input:["dog"],isUnique("dig"),isUnique("dug"),isUnique("dag"),isUnique("dog"),isUnique("doge")

Output:[true,true,true,true,true]

Expected:[false,false,false,true,true]

Input:[],isUnique("hello")

Output:[false]

Expected:[true]

Input:["hello"],isUnique("hello")

Output:[false]

Expected:[true]

Input:["a","a"],isUnique("a")

Output:[false]