LeetCode - 125. Valid Palindrome

Thoughts:

I choose two pointer to solve this problem, when I need to compare two values.

The time complexity is O(n) for the iteration.

Thoughts visuzlization:

s = "A man, a plan, a canal: Panama"

first, make all the characters lowercase
second, split the string into an array
third, transform characters into ascii code and filter the non-alphanumeric characters
fourth, join the array back into a string

then compare the two values by two pointer

"amanaplanacanalpanama"
 l                   r
"amanaplanacanalpanama"
  l                 r
"amanaplanacanalpanama"
   l               r
... and so on

if there is a mismatch, return false
and just return true in the end

Code:

/**
 * @param {string} s
 * @return {boolean}
 */
var isPalindrome = function (s) {
  const filtered = s
    .toLowerCase()
    .split('')
    .filter((char) => {
      const ascii = char.charCodeAt(0);
      const isNumber = ascii >= 48 && ascii <= 57;
      const isAlphabet = ascii >= 97 && ascii <= 122;
      return isNumber || isAlphabet;
    })
    .join('');

  let l = 0;
  let r = filtered.length - 1;
  while (l < r) {
    if (filtered[l] !== filtered[r]) {
      return false;
    }
    l++;
    r--;
  }
  return true;
};

Reference

Original link