Valid Palindrome

本文介绍了一种使用C++标准库函数实现字符串回文判断的方法。通过运用tolower转换所有字符为小写,并利用迭代器忽略非字母数字字符,实现了高效简洁的回文判断算法。

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遇到一些新的函数<algorithm>中的transform

class Solution {
public:
    bool isPalindrome(string s) {
        if (s.size() == 0)return true;
        transform(s.begin(), s.end(), s.begin(), ::tolower);
        string::iterator left = s.begin();
        string::iterator right = s.end();
        while(left < right){
            if(!isalnum(*left)){
                ++left;
                continue;
            }
            if(!isalnum(*right)){
                --right;
                continue;
            }
            if(*left == *right){
                ++left;
                --right;
            }else{
                return false;
            }
        }
        return true;
    }
};


OutputIterator transform ( InputIterator first1, InputIterator last1, OutputIterator result, UnaryOperator op );

isalnum()

tolower()

### 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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