leetCode#125. Valid Palindrome

本文主要是介绍leetCode#125. Valid Palindrome，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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) - 1while start <= end:while start <= end and not self.checkValid(s[start]):start += 1while start <= end and not self.checkValid(s[end]):end -= 1if start <= end and self.checkNotSame(ord(s[start]), ord(s[end])) and self.checkNotSame(ord(s[end]), ord(s[start])):return Falsestart, end = start + 1, end - 1return Truedef 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 != belse: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) - 1while start < end:while start < end and not s[start].isalnum():start += 1while start < end and not s[end].isalnum():end -= 1if s[start].lower() != s[end].lower():return Falsestart, end = start + 1, end - 1return 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。