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 <= 1001 <= 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.
博主正在刷题的时候，突然朋友圈刷出了科比坠机的消息，惊的下巴都掉了，忙看了下日期，不是四月一啊，于是疯狂的 google，中文搜不到任何相关的消息，于是搜英文 Kobe Bryant，结果真的有坠机消息，而且是几分钟前刚发布的，渐渐的很多微信群里都开始讨论了，连 wiki 上都更新了，随着越来越多的媒体确认这一个消息，心情越来越沉重了。算起来了，在博主最早关注 NBA 的时候，科比就当红球星，二十年的光辉岁月，五座总冠军戒指，甚至退役后还获得过奥斯卡小金人，年仅四十一岁，本来是要续写另一段传奇人生，就这么的走了？人生无常啊，你永远不知道意外和明天哪一个先到来，能平平安安的活着就已经是万幸了。RIP，一路走好，科比，愿天堂没有直升机。下面带着沉重的心情来做题吧，这道题是关于邮件的，邮件名里可能会有两个特殊符号，点和加号，对于点采取直接忽略的做法，对于加号则是忽略其后面所有的东西，现在问我们有多少个不同的邮箱。没有太多的技巧，就直接遍历一下所有的字符，遇到点直接跳过，遇到 '+' 或者 '@' 直接 break 掉。注意这里其实有个坑，就是域名中也可能有点，而这个点是不能忽略的，所以要把 '@' 及其后面的域名都提取出来，连到之前处理好的账号后面，一起放到一个 HashSet 中，利用其可以去重复的特性，最终剩余的个数即为所求，参见代码如下：
class Solution {
public:
int numUniqueEmails(vector<string>& emails) {
unordered_set<string> st;
for (string email : emails) {
string name;
for (char c : email) {
if (c == '.') continue;
if (c == '+' || c == '@') break;
name.push_back(c);
}
name += email.substr(email.find('@'));
st.insert(name);
}
return st.size();
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/929
参考资料：
https://leetcode.com/problems/unique-email-addresses/
https://leetcode.com/problems/unique-email-addresses/discuss/317207/C%2B%2B-Concise-Solution
https://leetcode.com/problems/unique-email-addresses/discuss/186798/Java-7-liner-with-comment.
[LeetCode All in One 题目讲解汇总(持续更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)