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In this post, you will find the solution for the **Palindrome Number** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Palindrome Number– LeetCode Problem

Palindrome Number– LeetCode Problem

**Problem:**

Given an integer `x`

, return `true`

if `x`

is palindrome integer.

An integer is a **palindrome** when it reads the same backward as forward.

- For example,
`121`

is a palindrome while`123`

is not.

**Example 1:**

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

**Example 2:**

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

**Example 3:**

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

**Constraints:**

`-2`

^{31}<= x <= 2^{31}- 1

Palindrome Number– LeetCode Solutions

class Solution { public: bool isPalindrome(int x) { if (x < 0) return false; long reversed = 0; int y = x; while (y) { reversed = reversed * 10 + y % 10; y /= 10; } return reversed == x; } };

class Solution { public boolean isPalindrome(int x) { if (x < 0) return false; long reversed = 0; int y = x; while (y > 0) { reversed = reversed * 10 + y % 10; y /= 10; } return reversed == x; } }

class Solution: def isPalindrome(self, x: int) -> bool: if x < 0: return False rev = 0 y = x while y: rev = rev * 10 + y % 10 y //= 10 return rev == x