Palindrome Number Check

Given an integer x, return True if x is a palindrome number, and False otherwise. Do NOT convert the integer to a string. Input: 121 → True. Input: -121 → False (reads as 121- backwards). Input: 10 → False (reads as 01).

Combine the digit extraction technique from Day 1 (Reverse Number) with the palindrome concept from Week 1. Reverse the entire number mathematically, then compare with the original. If they match, it is a palindrome. Negative numbers are never palindromes because the minus sign doesn't mirror.

1. If x < 0, return False immediately (negative numbers can't be palindromes). 2. Store the original value of x. 3. Build the reversed number using the digit extraction loop: digit = x % 10, reverse = reverse * 10 + digit, x = x // 10. 4. Compare the reversed number with the original. If equal, return True.

- Using str(x) to convert and compare — this violates the constraint - Forgetting that negative numbers are not palindromes - Not preserving the original value before modifying x in the loop - Edge case: 0 is a palindrome (it reads the same both ways)

This problem elegantly combines two concepts: the digit extraction pattern and the palindrome check. It shows how mathematical operations can replace string manipulation, often with better space efficiency. This is a classic LeetCode easy problem (Problem #9).