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Java-based LeetCode algorithm problem solutions, regularly updated
package g0001_0100.s0009_palindrome_number;
// #Easy #Math #Udemy_Integers #2023_08_09_Time_4_ms_(100.00%)_Space_43_MB_(43.42%)
/**
* 9 - Palindrome Number\.
*
* Easy
*
* 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 palindrome while `123` is not.
*
* **Example 1:**
*
* **Input:** x = 121
*
* **Output:** true
*
* **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.
*
* **Example 4:**
*
* **Input:** x = -101
*
* **Output:** false
*
* **Constraints:**
*
* * -231 <= x <= 231 - 1
*
* **Follow up:** Could you solve it without converting the integer to a string?
**/
public class Solution {
public boolean isPalindrome(int x) {
if (x < 0) {
return false;
}
int rev = 0;
int localX = x;
while (localX > 0) {
rev *= 10;
rev += localX % 10;
localX /= 10;
}
return rev == x;
}
}
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