class Solution(object):
  def isPalindrome(self, s):
    """
    :type s: str
    :rtype: bool
    """
    # 創(chuàng)建一個(gè)空字符串
    s_new = ''
    # 遍歷字符串s
    for i in s:
     # 判斷，如果是字母或數(shù)字，將其轉(zhuǎn)為小寫(xiě)拼接到字符串中
      if i.isalnum():
        s_new += i.lower()
    # 切片后s_new[::-1]與s_new比較，并將結(jié)果返回
    return s_new[::-1] == s_new

class Solution(object):
  def isPalindrome(self, s):
    """
    :type s: str
    :rtype: bool
    """
    low = 0
    high = len(s) - 1
    #在字符串為空或只有一個(gè)字符時(shí)，返回True
    if len(s) <= 1:
      return True
    # 設(shè)定low和high對(duì)比的條件
    while low < high:
     # 如果不是字母或數(shù)字，low往后移一位【low < high為必須條件，不然會(huì)造成索引越界】
      while not s[low].isalnum() and low < high:
        low += 1
      # 如果不是字母或數(shù)字，high往前移一位
      while not s[high].isalnum() and low < high:
        high -= 1
       # 判斷：如果相同，繼續(xù)下一次對(duì)比；如果不相同，直接返回False
      if s[low].lower() == s[high].lower():
        low += 1
        high -= 1
      else:
        return False
    # low和high重合，即退出循環(huán)，表示前后都是一一對(duì)應(yīng)的，返回True
   return True

def isPalindrome(self, s):
    """
    :type s: str
    :rtype: bool
    """
    s = s.lower()
    alphanumeric = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','0','1','2','3','4','5','6','7','8','9']
    newStr = ""
    for i in s:
      if i in alphanumeric:
        newStr += i
    return newStr==newStr[::-1]

def isPalindrome(self, s):
    """
    :type s: str
    :rtype: bool
    """
    s = s.lower()
    newStr = ""
    for i in s:
      if i.isalnum():
        newStr += i
    return newStr==newStr[::-1]

def isPalindrome(self, s):
    """
    :type s: str
    :rtype: bool
    """
    s = s.lower()
    import re
    s = re.sub('[^a-z0-9]', "", s)
    return s==s[::-1]