Every email consists of a local name and a domain name, separated by the @ sign.
For example, in?[email protected]
,?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,?"[email protected]"
?and?"[email protected]"
?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?[email protected]
?will be forwarded to?[email protected]
.? (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: ["[email protected]","[email protected]","[email protected]"]
Output: 2
Explanation:?"[email protected]" and "[email protected]" 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.
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...)
原文地址:https://www.cnblogs.com/grandyang/p/12235380.html