LeetCode 125. Valid Palindrome

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本文介绍了一个使用双指针法来判断字符串是否为回文的方法，只考虑字母和数字字符，并忽略大小写。通过具体的C++代码实现，展示了如何有效地解决这一常见编程问题。

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.

// two pointer problem

#include <string>
#include <iostream>
using namespace std;

bool isPalindrome(string s) {
    if(s.size() <= 1) return true;
    int i = 0;
    int j = s.size() - 1;
    while(i <= j) {
        if(isalnum(s[i]) && isalnum(s[j])) {
            if(tolower(s[i]) == tolower(s[j])) {
                i++;
                j--;
                continue;
            } else {
                return false;
            }
        }
        if(!isalnum(s[i])) {i++;}
        if(!isalnum(s[j])) {j--;}
    }
    return true;
}

int main(void) {
    string tmp = "A man, a plan, a canal : Panama six";
    bool res = isPalindrome(tmp);
    cout << res << endl;
}