leetcode125之验证回文串

题目描述：

给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写。

说明：本题中，我们将空字符串定义为有效的回文串。

示例 1:

输入: "A man, a plan, a canal: Panama"
输出: true

示例 2:

输入: "race a car"
输出: false

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-palindrome

代码实现：

def isPalindrome(s):
    '''
    验证是否为回文串
    :param str:
    :return:
    '''

    code = ["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"]
    CODE = ["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"]
    num = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
    string = ""
    for ch in s:
        if ch in code:
            string += ch
        elif ch in CODE:
            string += code[CODE.index(ch)]
        elif ch in num:
            string += ch

    print("string=", string)
    if string.isspace() or len(string) == 0:
        return True

    left, right = 0, len(string) - 1
    while left < right:
        if string[left] == string[right]:
            left += 1
            right -= 1
        else:
            return False

    return True


print("------测试isHuiwen--------")
s = "aaaa"
print(isPalindrome(s))


def isPalindrome1(s):
    '''
    反转字符串来判断
    :param s:
    :return:
    '''
    # isalnum() 方法检测字符串是否由字母和数字组成。
    sgood = "".join(ch.lower() for ch in s if ch.isalnum())
    return sgood == sgood[::-1]
    # return tmps == tmps[::-1]


print("------测试isHuiwen--------")
s = "A man, a plan, a canal: Panama"
print(isPalindrome1(s))


def isPalindromel2(s):
    '''
    双指针判断
    :param s:
    :return:
    '''
    tmps = ""
    for ch in s:
        if ch.isalnum():
            tmps += ch.lower()

    print("tmps=", tmps)
    left, right = 0, len(tmps) - 1
    while left < right:
        if tmps[left] == tmps[right]:
            left += 1
            right -= 1
        else:
            return False

    return True


print("------测试isHuiwen2--------")
s = "s "
print(isPalindromel2(s))


def isPalinedrome3(s):
    left, right = 0, len(s) - 1
    while left < right:
        if s[left].isalnum() == True and s[right].isalnum() == True:
            if s[left].lower() == s[right].lower():
                left += 1
                right -= 1
            else:
                return False
        elif s[left].isalnum() == True and s[right].isalnum() == False:
            right -= 1
        elif s[left].isalnum() == False and s[right].isalnum() == True:
            left += 1
        else:
            left += 1
            right -= 1

    return True


print("------测试isHuiwen3--------")
s = " "
print(isPalinedrome3(s))

输出：

------测试isHuiwen--------
string= aaaa
True
------测试isHuiwen--------
True
------测试isHuiwen2--------
tmps= s
True
------测试isHuiwen3--------
True

总结：双指针法和字符串反转直接判断两种方法。isalnum()函数判断字符是否为数字或者字母。字符串的反转操作[::-1]