当前位置：首页 > news >正文

LeetCode 831. Masking Personal Information【字符串,正则表达式】中等

news 2025/7/18 13:59:10

本文属于「征服LeetCode」系列文章之一，这一系列正式开始于2021/08/12。由于LeetCode上部分题目有锁，本系列将至少持续到刷完所有无锁题之日为止；由于LeetCode还在不断地创建新题，本系列的终止日期可能是永远。在这一系列刷题文章中，我不仅会讲解多种解题思路及其优化，还会用多种编程语言实现题解，涉及到通用解法时更将归纳总结出相应的算法模板。

为了方便在PC上运行调试、分享代码文件，我还建立了相关的仓库。在这一仓库中，你不仅可以看到LeetCode原题链接、题解代码、题解文章链接、同类题目归纳、通用解法总结等，还可以看到原题出现频率和相关企业等重要信息。如果有其他优选题解，还可以一同分享给他人。

由于本系列文章的内容随时可能发生更新变动，欢迎关注和收藏征服LeetCode系列文章目录一文以作备忘。

You are given a personal information string s, representing either an email address or a phone number. Return the masked personal information using the below rules.

Email address: An email address is:

A name consisting of uppercase and lowercase English letters, followed by
The '@' symbol, followed by
The domain consisting of uppercase and lowercase English letters with a dot '.' somewhere in the middle (not the first or last character).

To mask an email:

The uppercase letters in the name and domain must be converted to lowercase letters.
The middle letters of the name (i.e., all but the first and last letters) must be replaced by 5 asterisks "*****".

Phone number: A phone number is formatted as follows:

The phone number contains 10-13 digits.
The last 10 digits make up the local number.
The remaining 0-3 digits, in the beginning, make up the country code.
Separation characters from the set {'+', '-', '(', ')', ' '} separate the above digits in some way.

To mask a phone number:

Remove all separation characters.
The masked phone number should have the form:
- "***-***-XXXX" if the country code has 0 digits.
- "+*-***-***-XXXX" if the country code has 1 digit.
- "+**-***-***-XXXX" if the country code has 2 digits.
- "+***-***-***-XXXX" if the country code has 3 digits.
"XXXX" is the last 4 digits of the local number.

Example 1:

Input: s = "LeetCode@LeetCode.com"
Output: "l*****e@leetcode.com"
Explanation: s is an email address.
The name and domain are converted to lowercase, and the middle of the name is replaced by 5 asterisks.

Example 2:

Input: s = "AB@qq.com"
Output: "a*****b@qq.com"
Explanation: s is an email address.
The name and domain are converted to lowercase, and the middle of the name is replaced by 5 asterisks.
Note that even though "ab" is 2 characters, it still must have 5 asterisks in the middle.

Example 3:

Input: s = "1(234)567-890"
Output: "***-***-7890"
Explanation: s is a phone number.
There are 10 digits, so the local number is 10 digits and the country code is 0 digits.
Thus, the resulting masked number is "***-***-7890".

Constraints:

s is either a valid email or a phone number.
If s is an email:
- 8 <= s.length <= 40
- s consists of uppercase and lowercase English letters and exactly one '@' symbol and '.' symbol.
If s is a phone number:
- 10 <= s.length <= 20
- s consists of digits, spaces, and the symbols '(', ')', '-', and '+'.

题意：返回按照题目规则、隐藏个人信息后的电子邮件地址或电话号码。

解法模拟

首先判断 $s$ 是邮箱还是电话号码。显然，如果 $s$ 中有字符 @ ，那么它是邮箱，否则它是电话号码。

如果 $s$ 是邮箱，我们将 $s$ 的 @ 之前的部分保留第一个和最后一个字符，中间用 ***** 代替，并将整个字符串转换为小写。

如果 $s$ 是电话号码，我们只保留 $s$ 中的所有数字，用 regex_replace 将符合匹配规则的子串替换为其他字符串。再将最后 $10$ 位本地号码变成 ***-***-XXXX 的形式，并判断 $s$ 中是否有额外的国际号码。如果有，则将国际号码之前添加 + 号并加到本地号码的最前端，具体讨论如下：

如果有 $10$ 位数字，则加上前缀位空字符串。
如果有 $11$ 位数字，则加上前缀 +*- 。
如果有 $12$ 位数字，则不加上前缀 +**- 。
如果有 $13$ 位数字，则不加上前缀 +***- 。

class Solution {
private:vector<string> country = {"", "+*-", "+**-", "+***-"};
public:string maskPII(string s) {size_t idx = s.find('@'); // 有就是邮箱if (idx != string::npos) {transform(s.begin(), s.end(), s.begin(), ::tolower); // 全部小写return s.substr(0, 1) + "*****" + s.substr(idx - 1);// 名字第一个字母,中间字母用*****替换,名字最后一个字母和后面的域名不变}s = regex_replace(s, regex("[^0-9]"), ""); // 非数字字符替换为空return country[s.size() - 10] + "***-***-" + s.substr(s.size() - 4);}
};