LeetCode 929. Unique Email Addresses

929. Unique Email Addresses

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.

题目描述：大概是给你个邮箱的列表，给了一个 local name 的命名规则，问如果同时向这些邮箱中发送邮件，有多少不同邮箱能够实际收到这些邮件。

题目分析：挺水的一道题，首先，邮箱是由local name + @ + domain name 组成。我们从题意可以知道给出 local name 的两条命名规则：

遇到 . 直接无视。例如 alice.z@leetcode.com 和 alicez@leetcode.com 是等价的关系
遇到 + ，忽略 + 后面的内容。例如 m.y+name@email.com 和 my@email.com 是等价的关系

我们考虑用集合进行去重操作，同一邮箱收到多条邮件只能算做一次。枚举邮箱列表中的邮箱地址，根据 local name 的命名规则，通过 split 函数分割，将真实邮箱存入集合中，然后计算出的集合中的元素个数即为我们所求的邮箱数。

python 代码：

class Solution(object):
    def numUniqueEmails(self, emails):
        """
        :type emails: List[str]
        :rtype: int
        """
        real_email = set()
        for i in range(len(emails)):
            x = emails[i]
            s = str(x.split("+")[0].replace('.','') + '@' + x.split("@")[1])
            real_email.add(s)
        
        return len(real_email)

C++ 代码：

class Solution {
public:
    int numUniqueEmails(vector<string>& emails) {
        set<string> x;
        for(int i = 0; i < emails.size(); i++){
            if(emails[i].find('.') != string::npos){
                emails[i].erase(remove(emails[i].begin(), find(emails[i].begin(), emails[i].end(),'@'), '.'), find(emails[i].begin(), emails[i].end(), '@'));
            }
            if(emails[i].find('+') != string::npos){
                emails[i].erase(find(emails[i].begin(), find(emails[i].begin(), emails[i].end(), '@'), '+'), find(emails[i].begin(), emails[i].end(), '@'));
            }
            x.insert(emails[i]);
        }
        return x.size();
    }
};