Valid Palindrome II

Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.

Example 1:

Input: "aba"
Output: True

Example 2:

Input: "abca"
Output: True
Explanation: You could delete the character 'c'.

Note:

    1. The string will only contain lowercase characters a-z. The maximum length of the string is 50000.

 

这道题是验证回文字符,字符串中只含有小写字母,而且允许删除一个字符,那么当遇到不匹配的时候,这里是选择左边和右边都尝试一次删除,只要有一种能返回true,那么结果就返回true。


class Solution {
public:
    bool validPalindrome(string s) {
        int left = 0, right = s.size() - 1;
        while (left < right) {
            if (s[left] == s[right]) {
                ++left; --right;
            } else {
                int l = left, r = right - 1;
                while (l < r) {
                    if (s[l] != s[r]) break;
                    ++l; --r;
                    if (l >= r) return true;
                }
                ++left;
                while (left < right) {
                    if (s[left] != s[right]) return false;
                    ++left; --right;
                }
            }
        }
        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