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125. Valid Palindrome

Description

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 $i \geq j$:

  1. 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.
  2. 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.
  3. If the lowercase form of $s[i]$ and $s[j]$ are not equal, return false.
  4. 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 $O(1)$.

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class Solution:
    def isPalindrome(self, s: str) -> bool:
        i, j = 0, len(s) - 1
        while i < j:
            if not s[i].isalnum():
                i += 1
            elif not s[j].isalnum():
                j -= 1
            elif s[i].lower() != s[j].lower():
                return False
            else:
                i, j = i + 1, j - 1
        return True

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

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class Solution {
public:
    bool isPalindrome(string s) {
        int i = 0, j = s.size() - 1;
        while (i < j) {
            if (!isalnum(s[i])) {
                ++i;
            } else if (!isalnum(s[j])) {
                --j;
            } else if (tolower(s[i]) != tolower(s[j])) {
                return false;
            } else {
                ++i;
                --j;
            }
        }
        return true;
    }
};

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func isPalindrome(s string) bool {
    i, j := 0, len(s)-1
    for i < j {
        if !isalnum(s[i]) {
            i++
        } else if !isalnum(s[j]) {
            j--
        } else if tolower(s[i]) != tolower(s[j]) {
            return false
        } else {
            i, j = i+1, j-1
        }
    }
    return true
}

func isalnum(ch byte) bool {
    return (ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z') || (ch >= '0' && ch <= '9')
}

func tolower(ch byte) byte {
    if ch >= 'A' && ch <= 'Z' {
        return ch + 32
    }
    return ch
}

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function isPalindrome(s: string): boolean {
    let i = 0;
    let j = s.length - 1;
    while (i < j) {
        if (!/[a-zA-Z0-9]/.test(s[i])) {
            ++i;
        } else if (!/[a-zA-Z0-9]/.test(s[j]))