回文字符串判断python实现

def is_palindromic(num):
    str_len = len(num)
    index = str_len // 2
    str1 = num[:index]
    str2 = num[index:][::-1]
    if str1 == str2:
        return True
    else:
        return False

def is_palindromic2(num):
    count = 0
    for i in range(len(num)):
        count += 1
        if count < len(num) // 2:
            if num[i] != num[len(num) - i - 1]:
                return False
        else:
            break
    return True

def is_palindromic3(num):
    # 如果字符串只有0个或1个字符，那么该字符串符合回文的定义
    if len(num) < 2:
        return True
    # 如果字符串不止一个字符，那么检查字串符的第一项和最后一项是否等同
    if num[0] != num[-1]:  
        return False
    return is_palindromic3(num[1:-1])

def is_palindromic4(num):
    return num == num[::-1]