Leetcode 929. Unique email address

Link to the original question ：Leetcode 929. Unique Email Addresses

Every valid email consists of a local name and a domain name, separated by the @ sign. Besides lowercase letters, the email may contain one or more . or +.

For example, in [email protected], alice is the local name, and leetcode.com is the domain name.
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. Note that this rule does not apply to domain names.

For example, [email protected] and [email protected] forward to the same email address.
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. Note that this rule does not apply to domain names.

For example, [email protected] will be forwarded to [email protected].
It is possible to use both of these rules at the same time.

Given an array of strings emails where we send one email to each emails[i], return the number of different addresses that actually receive mails.

Example 1:

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

Example 2:

Input: emails = ["[email protected]","[email protected]","[email protected]"]
Output: 3

Constraints:

• 1 <= emails.length <= 100
• 1 <= emails[i].length <= 100
• emails[i] consist of lowercase English letters, ‘+’, ‘.’ and ‘@’.
• Each emails[i] contains exactly one ‘@’ character.
• All local and domain names are non-empty.
• Local names do not start with a ‘+’ character.
• Domain names end with the “.com” suffix.

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Method 1 ： Hashtable

Ideas ：

simulation , Convert your email address to a standard look . Through hash table unordered_set To save the converted email address , It can automatically remove the weight , Finally, return the length

c++ Code ：

class Solution {
    
public:
    int numUniqueEmails(vector<string>& emails) {
    
        unordered_set<string> st;

        for(auto email : emails){
    
            //  Local name part 
            string local;
            for(auto c : email){
    
                if(c == '+' || c == '@') break;
                if(c != '.') local += c;
            }
            st.emplace(local + email.substr(email.find('@')));
        }
        return st.size();
    }
};

Complexity analysis ：

• Time complexity ：O(n), You need to go through every character in the array ,n Is the sum of all character lengths
• Spatial complexity ：O(n), Save each character at most