leetCode#125. Valid Palindrome

Description

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.

Code

class Solution(object):
    def isPalindrome(self, s):
        """
        :type s: str
        :rtype: bool
        """
        start, end = 0, len(s) - 1
        while start <= end:
            while start <= end and not self.checkValid(s[start]):
                start += 1
            while start <= end and not self.checkValid(s[end]):
                end -= 1
            if start <= end and self.checkNotSame(ord(s[start]), ord(s[end])) and self.checkNotSame(ord(s[end]), ord(s[start])):
                return False
            start, end = start + 1, end - 1
        return True
    def checkValid(self, alpha):
        return (alpha >= '0' and alpha <= '9') or (alpha >= 'a' and alpha <= 'z') or (alpha >= 'A' and alpha <= 'Z')
    def checkNotSame(self, a, b):
        if (a >= 48 and a <= 57) or (b >= 48 and b <= 57):
            return a != b
        else:
            return a != b and a + 32 != b

里面有个小坑，当判断小写字母是否与大写字母相等时，偷懒写了个加32，结果测试用例里故意出了个”0P”这俩ascii刚好也相差32，恶意满满呀。于是得判断下a, b是否都为字母先。
写完后看了下别人的代码，原来python自带有isalnum()函数，就是用于判断是否字母和数字的…而且关于大小写的判断，可以直接将字符统一为大写或者小写就好了…所以精简的代码应该如下：

Code

class Solution(object):
    def isPalindrome(self, s):
        """
        :type s: str
        :rtype: bool
        """
        start, end = 0, len(s) - 1
        while start < end:
            while start < end and not s[start].isalnum():
                start += 1
            while start < end and not s[end].isalnum():
                end -= 1
            if s[start].lower() != s[end].lower():
                return False
            start, end = start + 1, end - 1
        return True

注意最外层的while循环条件可以改为小于号，等号可以去掉，因为此时当start == end时肯定字符也相等。同时里面的两个while也应该把等号去掉，如果加上了等号，会造成再次进循环，导致start > end（比如“.,”该用例），则不得不在最后一个if的时候还判断start <= end，造成冗余。

Conclusion

题目很简单，主要学习函数和小细节。