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UniqueEmailAddresses.cpp
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UniqueEmailAddresses.cpp
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// Source : https://leetcode.com/problems/unique-email-addresses/
// Author : Hao Chen
// Date : 2020-07-26
/*****************************************************************************************************
*
* 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,
* "[email protected]" and "[email protected]" forward to the same email address. (Note that
* this rule does not apply for domain names.)
*
* 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] will be forwarded to
* [email protected]. (Again, this rule does not apply for domain names.)
*
* 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:
* ]
* 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.
* All local and domain names are non-empty.
* Local names do not start with a '+' character.
*
******************************************************************************************************/
class Solution {
public:
int numUniqueEmails(vector<string>& emails) {
unordered_map<string, int> dict;
for (auto& email : emails) {
string addr;
for(int i=0; i<email.size(); i++) {
if (email[i] == '.' ) continue;
if (email[i] == '+') {
while( email[++i] != '@');
}
if (email[i] =='@') {
addr += email.substr(i);
break;
}
addr += email[i];
}
dict[addr]++;
}
return dict.size();
}
};