415. Valid Palindrome

本文介绍了一种使用 Python 实现的高效算法,该算法能够在 O(n) 时间复杂度内判断一个字符串是否为回文,同时不使用额外内存。通过忽略所有非字母数字字符并忽略大小写差异来确定字符串的有效回文特性。

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Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Example

"A man, a plan, a canal: Panama" is a palindrome.

"race a car" is not a palindrome.

Challenge

O(n) time without extra memory.

class Solution:
    """
    @param s: A string
    @return: Whether the string is a valid palindrome
    """
    def isPalindrome(self, s):
        start, end = 0, len(s)-1
        while start < end:
            while start < end and not s[start].isalpha() and not s[start].isdigit():
                start += 1
            while start < end and not s[end].isalpha() and not s[end].isdigit():
                end -= 1
            if start <  end and s[start].lower() != s[end].lower():
                return False
            start += 1
            end -= 1
        return True

### XTUOJ Perfect Palindrome Problem Analysis For the **Perfect Palindrome** problem on the XTUOJ platform, understanding palindromes and string manipulation algorithms plays a crucial role. A palindrome refers to a word, phrase, number, or other sequences of characters which reads the same backward as forward[^1]. The challenge typically involves checking whether a given string meets specific conditions to be considered a perfect palindrome. In many similar problems, preprocessing steps such as converting all letters into lowercase (or uppercase) can simplify subsequent checks by ensuring case insensitivity during comparison operations. Additionally, removing non-alphanumeric characters ensures that only relevant symbols participate in determining if the sequence forms a valid palindrome[^2]. To determine if a string is a perfect palindrome, one approach iterates from both ends towards the center while comparing corresponding elements until reaching the midpoint without encountering mismatches: ```python def is_perfect_palindrome(s): cleaned_string = ''.join(char.lower() for char in s if char.isalnum()) left_index = 0 right_index = len(cleaned_string) - 1 while left_index < right_index: if cleaned_string[left_index] != cleaned_string[right_index]: return False left_index += 1 right_index -= 1 return True ``` This function first creates `cleaned_string`, stripping away any irrelevant characters and normalizing cases. Then through iteration with two pointers moving inward simultaneously (`left_index` starting at position 0 and `right_index` initially set to the last index), comparisons occur between pairs of opposing positions within the processed input string. If every pair matches perfectly throughout this process, then the original string qualifies as a "perfect palindrome".
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