LeetCode刷题：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.

  解题思路：

1. 若邮箱中含有‘+’，先将‘+’和‘@’之间的字符删掉
2. 将新的邮箱中‘@’之前的‘.’删掉
3. 将最后的邮箱插入到无序容器unordered_set<string> result;

   解答：

   class Solution {
   public:
    int numUniqueEmails(vector<string>& emails) {
        unordered_set<string> result; //无序容器-C++ Primer P394；
        for(auto email = emails.begin(); email != emails.end(); ++email)
        {
            auto plus_pos = (*email).find('+');//获取‘+’的位置
            auto at_pos = (*email).find('@');
            if(plus_pos < at_pos)
            {
                auto len = at_pos - plus_pos;
                (*email).erase(plus_pos, len);//移除‘+’和‘@’中间的内容
            }
            auto dot_pos = (*email).find('.');
            at_pos = (*email).find('@');
            while(dot_pos < at_pos)
            {
                //删除所有‘@’前的‘.’
                (*email).erase(dot_pos, 1);
                dot_pos = (*email).find('.');
                at_pos--;
            }
            result.insert(*email);
        }
        return result.size();
    }
   };