原题：

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.

思路：

两指针，从头尾同时开始找，比较是否相同。排除其他符号。

代码是

class Solution:
    def isPalindrome(self, s):
        """
        :type s: str
        :rtype: bool
        """
        if not s:
            return True
        strlist = s.lower()
        i = 0
        j = len(s) - 1
        letter_number = set(['1','2','3','4','5','6','7','8','9','0','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'])
        while i < j:
            # print('----')
            # print(i)
            # print(strlist[i])
            # print(j)
            # print(strlist[j])
            if strlist[i] in letter_number and strlist[j] in letter_number:
                if strlist[i] != strlist[j]:
                    return False
                else:
                    i += 1
                    j -= 1
            
            elif strlist[i] in letter_number and strlist[j] not in letter_number:
                j -= 1
            else:
                i += 1
        return True

别人的代码用了一个API， isalnum()

def isPalindrome(self, s):
    l, r = 0, len(s)-1
    while l < r:
        while l < r and not s[l].isalnum():
            l += 1
        while l <r and not s[r].isalnum():
            r -= 1
        if s[l].lower() != s[r].lower():
            return False
        l +=1; r -= 1
    return True

插播两个找到所有字母的方法：

import string
wList = []
for word in string.lowercase:
    wList.append(word)
print wList
 
#['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']

 list(map(chr, range(ord('a'), ord('z') + 1)))
#['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']
 [chr(x) for x in range(ord('a'), ord('z') + 1)]
#['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']

Screen Shot 2017-11-25 at 11.21.29 AM.png

Screen Shot 2017-11-25 at 11.26.51 AM.png