125 Valid Palindrome

本文介绍了一种有效的算法来判断一个字符串是否为回文,仅考虑字母和数字字符,并忽略大小写差异。通过双指针技术从两端向中间扫描,跳过非字母数字字符进行比较。

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题目链接:https://leetcode.com/problems/valid-palindrome/

题目:

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

For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

解题思路:
这道题对我来说意义很有意义,因为它是我暑假面试中遇到的一道题!!!
另外一道是 1 Two Sum 如此看来当时面试的几道题基本都来自 leetcode !!!
因为当时面试官给出了解答,所以至今记得解题思路。

  • 这道题只考虑数字和字母(字母有大小写之分),所以遇到其它符号时,自动跳过。
  • 具体来说就是,分别从字符串两边向中间遍历,遇到非数字和字母的符号就跳过,直至左右两个指针遇到。

这题有些隐藏的小细节需要注意。
1. 字母有大小写之分,比较之前应该先统一转变为小写或大写。
2. 如果输入的字符串没有数字或字母,该字符串仍然看做是回文。
3. 在指针向中间靠拢的过程中,若指针一直没有遇到数字或字母,就会一直加加或减减,此时应注意判断指针是否越界。

代码实现:

public class Solution {
    public boolean isPalindrome(String s) {
        if(s == null || s.length() == 0 || s.length() == 1)
            return true;
        int l = 0;
        int r = s.length() - 1;
        while(l < r) {
            char lc = s.charAt(l);
            while((lc < '0' || lc > '9') && (lc < 'A' || lc > 'Z') && (lc < 'a' || lc > 'z')) {
                if(l >= r)
                    break;
                l ++;
                lc = s.charAt(l);
            }
            char rc = s.charAt(r);
            while((rc < '0' || rc > '9') && (rc < 'A' || rc > 'Z') && (rc < 'a' || rc > 'z')) {
                if(l >= r)
                    break;
                r --;
                rc = s.charAt(r);
            }
            if(Character.toLowerCase(s.charAt(l)) != Character.toLowerCase(s.charAt(r)))
                return false;
            l ++;
            r --;
        }
        return true;
    }
}
475 / 475 test cases passed.
Status: Accepted
Runtime: 9 ms
### 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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