Valid Palindrome

本文介绍了一种方法来判断给定的字符串是否为回文,特别考虑了包含标点符号和不同大小写的情况。通过使用Java代码实现,首先去除所有非字母数字字符并统一大小写,然后检查处理后的字符串是否与其反转相同。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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.

判断一个字符串是否为回文字符串,字符串中包含标点还有大小写的字母,我们可以先将字符串转换一下,用replaceAll方法去掉不用比较的字符,如何保留需要比较的字符我们用到正则表达式,处理之后在进行判断。代码如下:

public class Solution {
public boolean isPalindrome(String s) {
if(s == null) return true;
String ss = s.replaceAll("[^0-9a-zA-Z]","").toLowerCase();
return isPalin(ss);
}
public boolean isPalin(String s) {
int l = 0;
int r = s.length() - 1;
while(l < r) {
if(s.charAt(l) != s.charAt(r))
return false;
l ++;
r --;
}
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".
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值