125. Valid Palindrome(双指针)

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.

 

class Solution:
    def isPalindrome(self, s: str) -> bool:
        n = len(s)
        left, right = 0, n - 1
        
        while left < right:
            while left < right and not s[left].isalnum():
                left += 1
            while left < right and not s[right].isalnum():
                right -= 1
            if left < right:
                if s[left].lower() != s[right].lower():
                    return False
                left, right = left + 1, right - 1

        return True

 

 

 

 

class Solution {
public:
    bool is_ok(char c) {
        return (('0' <= c && c <= '9')  ||  ('a' <= c && c  <='z') || ('A' <= c && c  <='Z'));
    }
    bool isPalindrome(string s) {
        int l = 0, r = s.size() -1;
        while(l < r) {
            while(l < r && !is_ok(s[l]) ) l++;
            while(l < r && !is_ok(s[r]) ) r--;

            if (tolower(s[l]) != tolower(s[r])) return false;
            l++;r--;
        }
        return true;
    }
};

 

 

 

 

 

class Solution:
    def isPalindrome(self, s):
        """
        :type s: str
        :rtype: bool
        """
        if s=='':
            return True
        import re
        x = re.sub('\W','',s).lower()
        if x=='':
            return True
        if(len(x)%2!=0):#数
            for i in range(int(len(x)/2)+1):
                if(x[i]!=x[len(x)-i-1]):
                    return False
        else:
            for i in range(int(len(x)/2)+1):
                if(x[i]!=x[len(x)-i-1]):
                    return False
        return True