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题目：
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.
链接： http://leetcode.com/problems/valid-palindrome/
一刷，用两个指针从两个方向，出现的问题
1. string methods掌握不熟，str.isalpha(), str.isdigit(), str.isalnum()
2. 内部两个while loop之后需要再检查index是否valid，否则或者应该跳到外循环检查
3. 不要忘记更新下标
class Solution(object): def isPalindrome(self, s): if not s or s == ‘‘: return True i, j = 0, len(s) - 1 while i <= j: while i < len(s) and not s[i].isalnum(): i += 1 while j >= 0 and not s[j].isalnum(): j -= 1 if i == len(s) or j < 0: break if s[i].isalpha() and s[j].isalpha() and s[i].lower() == s[j].lower() or s[i].isdigit() and s[j].isdigit() and s[i] == s[j]: pass else: return False i += 1 j -= 1 return True
对比相同方法别人的代码，可以有以下改动：
class Solution(object): def isPalindrome(self, s): i, j = 0, len(s) - 1 while i <= j: while i < j and not s[i].isalnum(): i += 1 while i < j and not s[j].isalnum(): j -= 1 if s[i].lower() != s[j].lower(): return False i += 1 j -= 1 return True
别人还有更pythonic但是需要额外空间复杂度的做法，很赞，平时工作完全能写得出，但是刷题时就忘记了：
class Solution(object): def isPalindrome(self, s): cleanlist = [c for c in s.lower() if c.isalnum()] return cleanlist == cleanlist[::-1]
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原文地址：http://www.cnblogs.com/panini/p/5615978.html
