# 929 Unique Email Addresses

Regex 還不熟，需要再複習一下

LeetCode

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

For example, in [e-mail], 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, "[e-mail]" and "[e-mail]" 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 [e-mail] will be forwarded to [e-mail].  
(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? 

input: 給一有眾多 email(除了字母還會有 . 跟 +) 的陣列 
解釋: email 是 local name +  domain name 組成
若 local name 有 . 則 ignore
若 local name 有 + 則後面字母全部 ignore
output: 到底會真正寄出幾個 valid email

Example 1:

Input: ["[e-mail]",
"[e-mail]",
"[e-mail]"]

Output: 2
Explanation: "[e-mail]" and "[e-mail]" 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.

/**
 * @param {string[]} emails
 * @return {number}
 */
var numUniqueEmails = function(emails) {
    
};

怎麼解

會用 regex 篩選

• 抓 @ 前面但不包和 + 後面的字

• 抓 domain name

str.replace(/([^+@]*)\+?.*@(.*)/, newSubstr)

• ignore .

function newSubstr (match, p1, p2){
  return p1.replace(/\./g, '')+ '@' + p2;
}

function newSubstr (match, p1, p2){
  return p1.replace(/\./g, '')+ '@' + p2;
}
var numUniqueEmails = function(emails) {
 let newSet = new Set()
  emails.forEach( email => {
    newSet.add(email.replace(/([^+@]*)\+?.*@(.*)/, newSubstr))
  })
 return newSet.size;
};

學到什麼?

Regex 還不熟，需要複習一下