LeetCode-9 Palindrome Number

9. Palindrome Number easy

Given an integer x, return true if x is a 

, and false otherwise.

Example 1:

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Example 2:

Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Constraints:

• -231 <= x <= 231 - 1

Follow up: Could you solve it without converting the integer to a string?

Solutions:

a. Convert  Integer to String:

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

or

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

b. Dont't Convert  Integer to String:

class Solution:
    # This function checks if a given integer is a palindrome.
    def isPalindrome(self, x: int) -> bool:
        # Check if the number is negative or a multiple of 10 (excluding 0)
        if x < 0 or (x != 0 and x % 10 == 0):
            return False

        reversed_num = 0
        original = x

        # Reverse the second half of the number
        while x > reversed_num:
            reversed_num = reversed_num * 10 + x % 10  # Add the last digit of 'x' to 'reversed_num'
            x //= 10  # Remove the last digit of 'x'

        # Check if the reversed number matches the original number, x == reversed_num // 10  is when X have odd width
        return x == reversed_num or x == reversed_num // 10