2299. Strong Password Checker II

仲孙俊贤
2023-12-01

题目

A password is said to be strong if it satisfies all the following criteria:

It has at least 8 characters.
It contains at least one lowercase letter.
It contains at least one uppercase letter.
It contains at least one digit.
It contains at least one special character. The special characters are the characters in the following string: “!@#$%^&*()-+”.
It does not contain 2 of the same character in adjacent positions (i.e., “aab” violates this condition, but “aba” does not).
Given a string password, return true if it is a strong password. Otherwise, return false.

Example 1:

Input: password = "IloveLe3tcode!"
Output: true
Explanation: The password meets all the requirements. Therefore, we return true.

Example 2:

Input: password = "Me+You--IsMyDream"
Output: false
Explanation: The password does not contain a digit and also contains 2 of the same character in adjacent positions. Therefore, we return false.

Example 3:

Input: password = "1aB!"
Output: false
Explanation: The password does not meet the length requirement. Therefore, we return false.

Constraints:

1 <= password.length <= 100
password consists of letters, digits, and special characters: "!@#$%^&*()-+".

解题思路

简单题,注意大小写可以用py自带的lower()和upper()来判断是否和原字符串相等,另外set化string会直接返回set of character

代码

class Solution:
    def strongPasswordCheckerII(self, password: str) -> bool:
        is_strong_password = True
        # lengh checker
        if len(password) < 8:
            is_strong_password = False
        # lowercase / uppercase
        if password == password.lower() or password == password.upper():
            is_strong_password = False
        # digit
        if not set(map(str, range(0, 10))) & set(password):
            is_strong_password = False
        # special character
        if not set('!@#$%^&*()-+') & set(password):
            is_strong_password = False
        # consective character
        for i in range(len(password) - 1):
            if password[i] == password[i + 1]:
                is_strong_password = False
                break
        return is_strong_password

另外还看到了用掩码来判断各个状态的方法,感觉从代码角度更高效一些

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