Solutions

1. Convert to String

class Solution:
    def isPalindrome(self, x: int) -> bool:
        s = str(x)
        return s == s[::-1]

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

Where $n$ is the number of digits in the given integer.

2. Convert to String (Optimal)

class Solution:
    def isPalindrome(self, x: int) -> bool:
        s = str(x)
        n = len(s)
        for i in range(n // 2):
            if s[i] != s[n - i - 1]:
                return False
        return True

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

Where $n$ is the number of digits in the given integer.

3. Reverse the Integer

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False

        rev = 0
        num = x
        while num:
            rev = (rev * 10) + (num % 10)
            num //= 10

        return rev == x

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

Where $n$ is the number of digits in the given integer.

4. Two Pointers

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False

        div = 1
        while x >= 10 * div:
            div *= 10

        while x:
            if x // div != x % 10:
                return False
            x = (x % div) // 10
            div //= 100

        return True

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

Where $n$ is the number of digits in the given integer.

5. Reverse Half of the Number

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0 or (x != 0 and x % 10 == 0):
            return False

        rev = 0
        while x > rev:
            rev = (rev * 10) + (x % 10)
            x //= 10

        return x == rev or x == rev // 10

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

Where $n$ is the number of digits in the given integer.