leetcode 2299. Strong Password Checker II （python）

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描述

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.
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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.
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Example 3:

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

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

根据题意，如果满足以下所有条件，则称密码为强密码：

• 它至少有 8 个字符。
• 它至少包含一个小写字母。
• 它至少包含一个大写字母。
• 它至少包含一位数字。
• 它至少包含一个特殊字符。 特殊字符是以下字符串中的字符：“!@#$%^&*()-+”。
• 它在相邻位置不包含 2 个相同的字符（即，“aab”违反了此条件，但“aba”没有）。

给定一个字符串密码，如果它是一个强密码，则返回 true。 否则，返回 False 。

这道题其实就是考查对字符串的操作，最简单的就是使用正则进行解题，我直接用了最暴力的方法，不用费脑子。

时间复杂度为 O(N) ，空间复杂度为 O(N) 。

解答

class Solution(object):
    def strongPasswordCheckerII(self, password):
        """
        :type password: str
        :rtype: bool
        """
        counter = collections.Counter(list(password))
        keys = set(counter.keys())
        if len(password) < 8:
            return False
        if not keys & set(list('qwertyuiopasdfghjklzxcvbnm')):
            return False
        if not keys & set(list('QWERTYUIOPASDFGHJKLZXCVBNM')):
            return False
        if not keys & set(list('1234567890')):
            return False
        if not keys & set(list('!@#$%^&*()-+')):
            return False
        for i in range(1, len(password)):
            if password[i-1] == password[i]:
                return False
        return True
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运行结果

148 / 148 test cases passed.
Status: Accepted
Runtime: 17 ms
Memory Usage: 13.5 MB
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原题链接

leetcode.com/contest/biw…

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