125驗證迴文串

# 利用雙指針，一次遍歷，求出結果class Solution:    def isPalindrome(self, s: str) -> bool:        # 定義變量，接收字符串的長度        length = len(s)        # 長度小於等於1直接返回真        if length <= 1:return True        # 定義兩個指針， 分別指向字符串頭和尾        index1,index2 = 0,length - 1        while index1 <= index2:            # 判斷字符是否爲字母或者數字            if not s[index1].isalnum():                index1 += 1                continue            if not s[index2].isalnum():                index2 -= 1                continue            # 判斷兩個字符是否相同            if s[index1].lower() != s[index2].lower():                return False            index1 += 1            index2 -= 1        return TrueA = Solution()print(A.isPalindrome("A man, a plan, a canal: Panama"))print(A.isPalindrome(""))print(A.isPalindrome("qq"))print(A.isPalindrome("race a car"))