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 1:输入: "A man, a plan, a canal: Panama"输出: true示例 2:输入: "race a car"输出: false

/* * @lc app=leetcode id=125 lang=javascript * * [125] Valid Palindrome */// 只处理英文字符（题目忽略大小写，我们前面全部转化成了小写， 因此这里我们只判断小写）和数字function isValid(c) {  const charCode = c.charCodeAt(0);  const isDigit =    charCode >= "0".charCodeAt(0) && charCode <= "9".charCodeAt(0);  const isChar = charCode >= "a".charCodeAt(0) && charCode <= "z".charCodeAt(0);  return isDigit || isChar;}/** * @param {string} s * @return {boolean} */var isPalindrome = function (s) {  s = s.toLowerCase();  let left = 0;  let right = s.length - 1;  while (left < right) {    if (!isValid(s[left])) {      left++;      continue;    }    if (!isValid(s[right])) {      right--;      continue;    }    if (s[left] === s[right]) {      left++;      right--;    } else {      break;    }  }  return right <= left;};

class Solution {public:    bool isPalindrome(string s) {        if (s.empty())            return true;        const char* s1 = s.c_str();        const char* e = s1 + s.length() - 1;        while (e > s1) {            if (!isalnum(*s1)) {++s1; continue;}            if (!isalnum(*e)) {--e; continue;}            if (tolower(*s1) != tolower(*e)) return false;            else {--e; ++s1;}        }        return true;    }};

class Solution:    def isPalindrome(self, s: str) -> bool:        left, right = 0, len(s) - 1        while left < right:            if not s[left].isalnum():                left += 1                continue            if not s[right].isalnum():                right -= 1                continue            if s[left].lower() == s[right].lower():                left += 1                right -= 1            else:                break        return right <= left    def isPalindrome2(self, s: str) -> bool:        """        使用语言特性进行求解        """        s = ''.join(i for i in s if i.isalnum()).lower()        return s == s[::-1]

class Solution {    public boolean isPalindrome(String s) {        int n = s.length();        int left = 0, right = n - 1;        while (left < right) {            while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {                ++left;            }            while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {                --right;            }            if (left < right) {                if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {                    return false;                }                ++left;                --right;            }        }        return true;    }}