Description

Every email consists of a local name and a domain name, separated by the @ sign.

For example, in [email protected], alice is the local name, and leetcode.com is the domain name.

Besides lowercase letters, these emails may contain '.'s or '+'s.

If you add periods ('.') between some characters in the local name part of an email address, mail sent there will be forwarded to the same address without dots in the local name. For example, "

If you add a plus ('+') in the local name, everything after the first plus sign will be ignored. This allows certain emails to be filtered, for example [email protected]

It is possible to use both of these rules at the same time.

Given a list of emails, we send one email to each address in the list. How many different addresses actually receive mails?

Example 1:

Input: ["
 
[email protected]","[email protected]","[email protected]"]
Output: 2
Explanation: "[email protected]" and "[email protected]" actually receive mails

Note:

• 1 <= emails[i].length <= 100
• 1 <= emails.length <= 100
• Each emails[i] contains exactly one '@' character.

Analyse

簡單題，郵箱地址@前面的部分中的.直接省略，+後面的部分也省略, @後面的域名部分不作變換，剩下的就是真正的郵箱地址

[email protected] -> [email protected]

把變換後的郵箱地址存到set裡去就能拿到不同的郵箱地址數了

Code

第一個版本的程式碼，先把郵箱地址按@分割成兩部分，前面是localName，後面是domain，作完變換後insert到unordered_set中，輸出長度

48ms faster than 4.89%

int numUniqueEmails(vector<string>& emails)
{
    unordered_set<string> ss;
    string localName;
    string domain;
    string tmp = "";
    for(int i = 0; i < emails.size(); i++)
    {
        tmp = emails[i];
        size_t at_pos = tmp.find('@');
        if (at_pos != string::npos)
        {
            localName = tmp.substr(0, at_pos);
            domain = tmp.substr(at_pos + 1, emails[i].size());

            localName.erase(std::remove(localName.begin(), localName.end(), '.'), localName.end()); //去除localName中所有`.`

            size_t plus_pos = localName.find('+');
            localName = localName.substr(0, plus_pos);

            cout << localName + domain << endl;
            ss.insert(localName + domain);
        }
        else
        {
            continue;
        }
    }

    return ss.size();
}

上面那個版本的結果太差了，我猜測是erase那段花的時間太長了，自己用for迴圈實現了一遍，結果結果並沒有發生變化，還是48ms，直到我刪除了那行cout

20ms faster than 96.49%

int numUniqueEmails(vector<string>& emails) {
        unordered_set<string> ss;
        string email = "";
        for(int i = 0; i < emails.size(); i++)
        {
            email = emails[i];
            string address = "";
            size_t at_pos = email.find('@');
            string domain = email.substr(at_pos + 1, email.size());
            string localName = email.substr(0, at_pos);

            for(int j = 0; j < localName.size(); j++)
            {
                if (localName[j] == '+')
                {
                    break;
                }

                if (localName[j] == '.')
                {
                    continue;
                }

                address += localName[j];
            }

            ss.insert(address + domain);
        }

        return ss.size();
    }