[LeetCode] 929. Unique Email Addresses 唯一的电邮地址

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

For example, in alice@leetcode.com, 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, "alice.z@leetcode.com" and "alicez@leetcode.com"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 m.y+name@email.com will be forwarded to my@email.com.  (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: ["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]
Output: 2
Explanation: "testemail@leetcode.com" and "testemail@lee.tcode.com" 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.

 解法：根据题目要求，直接在@处分解email，把local name中的'.'和'+'之后的字符去掉，在和domain拼接成email地址，然后存进set，set可以去重，最后返回set的长度。

Java:

public int numUniqueEmails(String[] emails) {
        Set<String> normalized = new HashSet<>(); // used to save simplified email address, cost O(n) sapce.
        for (String email : emails) {
            String[] parts = email.split("@"); // split into local and domain parts.
            String[] local = parts[0].split("\+"); // split local by '+'.
            normalized.add(local[0].replace(".", "") + "@" + parts[1]); // remove all '.', and concatenate '@' and domain.        
        }
        return normalized.size();
    }　　

Python:

class Solution:
    def numUniqueEmails(self, emails):
        """
        :type emails: List[str]
        :rtype: int
        """
        email_set = set()
        for email in emails:
            local_name,domain_name = email.split("@")
            local_name ="".join(local_name.split('+')[0].split('.'))
            email = local_name +'@' + domain_name
            email_set.add(email)
        return len(email_set)

Python: wo

class Solution(object):
    def numUniqueEmails(self, emails):
        """
        :type emails: List[str]
        :rtype: int
        """
        ans = set()
        for email in emails:
            split_names = email.split('@')
            local_name = split_names[0]
            domain_name = split_names[1]
            local_name_remove_plus = local_name.split('+')[0]
            local_name_split_dot = local_name_remove_plus.split('.')
            local_name = ('').join(local_name_split_dot)
            new_email = local_name + '@' + domain_name
            ans.add(new_email)
        return len(ans)

C++:

class Solution
{
public:

    using Emails = vector< string >;
    using Unique = unordered_set< string >;

    int numUniqueEmails( Emails& emails, Unique unique={} )
    {
        for( auto& e: emails )
        {
            auto pivot = e.find_first_of( '@' );
            auto name = e.substr( 0, pivot ),
                 domain = e.substr( pivot );
            name.erase( remove( name.begin(), name.end(), '.' ), name.end() );
            auto pos = name.find_first_of( '+' );
            unique.insert( ( pos != string::npos )? name.erase( pos ) + domain : name + domain );
        }
        return static_cast< int >( unique.size() );
    }

};

　　

　　

　　

All LeetCode Questions List 题目汇总