A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string s, return true if it is a palindrome, or false otherwise.
Example 1:
Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.
Example 2:
Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.
Example 3:
Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.
Constraints:
1 <= s.length <= 2 * 105
s consists only of printable ASCII characters.
Solutions
Solution 1: Two Pointers
We use two pointers $i$ and $j$ to point to the two ends of the string $s$, and then loop through the following process until $i \geq j$:
If $s[i]$ is not a letter or a number, move the pointer $i$ one step to the right and continue to the next loop.
If $s[j]$ is not a letter or a number, move the pointer $j$ one step to the left and continue to the next loop.
If the lowercase form of $s[i]$ and $s[j]$ are not equal, return false.
Otherwise, move the pointer $i$ one step to the right and the pointer $j$ one step to the left, and continue to the next loop.
At the end of the loop, return true.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.