125. Valid Palindrome

Description

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

Note: For the purpose of this problem, we define empty string as valid palindrome.

Tags: Two Pointers、 String

思路 1

1 空串为true 2 从两边往中间靠拢, 遇到不是字母也不是数字的就跳过 3 左边游标索引大于右边游标索引则结束遍历 4 遍历过程中一旦发生不相等, 则不是回文 kotlin(228ms/94.44%):

class Solution {
    fun isPalindrome(s: String): Boolean {
        if (s == " ") return true
        val len = s.length
        var i = 0
        var j = len - 1
        while (i <= j) {
            try {
                while (!(s[i] in 'a'..'z' || s[i] in 'A'..'Z' || s[i] in '0'..'9')) i++
                while (!(s[j] in 'a'..'z' || s[j] in 'A'..'Z' || s[j] in '0'..'9')) j--
            } catch (e: Exception) {
                return true
            }
            try {
                if (!s[i++].equals(s[j--], true)) return false
            } catch (e: Exception) {
                return false
            }
        }
        return true
    }
}

JavaScript：

var isPalindrome = function(s) {
  var str = s.toLowerCase()
  let str1 = []
  for(var i = 0; i < str.length; i++) {
    if(/[a-z0-9]/ig.test(str[i])) {
        str1.push(str[i])
    }
  }
  return str1.join('') === str1.reverse().join('')
};

结语

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