125. Valid Palindrome

problem

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.

判断一个str是不是回文（无效字符忽略）

难度不高，注意的是空字符 and 数字也算有效字符

• 效率对比：30000的回文字符判断1000次

最初的解法:11.31s

discuss1解法:8.18s

discuss2解法：5.25s

1. wille循环改为strlist == strlist[::-1] 提升1s

2. str.isalnum(object) 比 object.isalnum()要慢

3. str(object)的转换时题目理解错误时使用isalpha需要的，现在冗余

修改下面后达到5.25s
strlist = "".join([x.upper() for x in s if x.isalnum()])

solution

class Solution(object):
    def isPalindrome(self, s):
        """
        :type s: str
        :rtype: bool
        """
        strlist = [x.upper() for x in s if str.isalnum(str(x))]
        if strlist:
            length = len(strlist)/2
            i = 0
            while i <= length:
                if strlist[i] != strlist[-i-1]:
                    return False
                i += 1
        return True

        

discuss1

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

discuss2

class Solution:
    # @param s, a string
    # @return a boolean
    def isPalindrome(self, s):
        s = "".join([c.lower() for c in s if c.isalnum()])
        return s == s[::-1]