Q929 Unique Email Addresses

Unique Email Addresses

Question

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.
• All local and domain names are non-empty.
• Local names do not start with a '+' character.

Approach 1: Split Words (Original Method)

Intuition and Algorithm

Note that split is a useful method.

class Solution(object):
    def reverseWords(self, s):
        out = []
        singleWord=s.split(' ')
        for iter in singleWord:
            iter=iter[::-1]# 反向排序数组
            out.append(iter)
        return ' '.join(out) # 返回字符串，用空格填充out中元素的间隔

def main():
    s="Let's take LeetCode contest"
    solution=Solution().reverseWords(s)
    print(solution)


if __name__ == "__main__":
    main()

simplification

class Solution(object):
    def reverseWords(self, s):
        return " ".join([i[::-1] for i in s.split()])

Complexity Analysis

• Time Complexity: O(n), where n is the total words of s.
• Space Complexity: O(1).
• Runtime: 20 ms, faster than 94.03% of Python online submissions for Reverse Words in a String III.
• Memory Usage: 13 MB, less than 43.54% of Python online submissions for Reverse Words in a String III.