[LeetCode] 125. 验证回文串

Posted 2022-04-05 powercai

题目链接 : https://leetcode-cn.com/problems/valid-palindrome/

题目描述:

给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写。

说明：本题中，我们将空字符串定义为有效的回文串。

示例:

示例 1:

输入: "A man, a plan, a canal: Panama"
输出: true

示例 2:

输入: "race a car"
输出: false

思路:

思路一: 用正则提取字母和数字字符, 在取反比较

思路二: 双指针

代码:

思路一:

def isPalindrome(self, s: str) -> bool:
        tmp = re.sub(r"[^A-Za-z0-9]","", s).lower()
        return tmp == tmp[::-1]

java

class Solution 
    public boolean isPalindrome(String s) 
        String tmp = s.replaceAll("[^A-Za-z0-9]", "").toLowerCase();
        String rev_tmp = new StringBuffer(tmp).reverse().toString();
        return tmp.equals(rev_tmp);
    

思路二:

def isPalindrome(self, s: str) -> bool:
        n = len(s)
        left = 0
        right = 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 s[left].lower() != s[right].lower():
                return False
            left += 1
            right -= 1
        return True

java

class Solution 
    public boolean isPalindrome(String s) 
        char[] c = s.toCharArray();
        int left = 0;
        int right = c.length - 1;
        while (left < right) 
            while (left < right && !Character.isLetterOrDigit(c[left])) left++;
            while (left < right && !Character.isLetterOrDigit(c[right])) right--;
            if (Character.toLowerCase(c[left]) != Character.toLowerCase(c[right])) return false;
            left++;
            right--;
        
        return true;