LeetCode: Valid Palindrome

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.

---

If the size of the string is even, the start checking position will be the middle two chars. Else if it is odd, the start checking position is the middle char.

C++ Code:

/*
 * func: is_palindrome
 * goal: to test if the input string is a palindrome
 * @param s: input string
 * return: true or false
 */
/*
 * find the symmetry axis and then validate
 * complexity: time O(n), space O(n)
 */
bool is_palindrome(string s){
    string convert = "";
    //remove non-alphanumeric chars
    for(const char &ch : s){
        if(isalnum(ch)){
            convert += isupper(ch) ? (ch + 32) : ch;
        }
    }
    int s_length = convert.length();
    if(s_length <= 1){
        return true;
    }
    //Find symmetry axis
    int mid_left = 0;
    int mid_right = 0;
    if(s_length & 1){
        mid_left = s_length/2;
        mid_right = s_length/2;
    }else{
        mid_left = s_length/2 - 1;
        mid_right = s_length/2;
    }
    while(mid_left >= 0 && mid_right <s_length){
        if(convert[mid_left] == convert[mid_right]){
            --mid_left;
            ++mid_right;
        }else{
            return false;
        }
    }
    return true;
}

Python Code:

# func: to determine if the input string is a palindrome
# @param s: input string
# @return: True or False
def is_palindrome(s):
    str = ''.join(x.lower() for x in s if x.isalnum())
    if len(str) <= 1:
        return True
    if len(str) & 1:
        mid_left = len(str)/2
        mid_right = len(str)/2
    else:
        mid_left = len(str)/2 - 1
        mid_right = mid_left + 1

    while mid_left >= 0 and mid_right < len(str):
        if str[mid_left] == str[mid_right]:
            mid_left -= 1
            mid_right += 1
        else:
            return False
    return True