# 9. Palindrome Number

## Description

Given an integer x, return true if x is a palindrome, 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

Solution 1: Reverse Half of the Number

First, we determine special cases:

• If $x < 0$, then $x$ is not a palindrome, directly return false;
• If $x > 0$ and the last digit of $x$ is $0$, then $x$ is not a palindrome, directly return false;
• If the last digit of $x$ is not $0$, then $x$ might be a palindrome, continue the following steps.

We reverse the second half of $x$ and compare it with the first half. If they are equal, then $x$ is a palindrome, otherwise, $x$ is not a palindrome.

For example, for $x = 1221$, we can reverse the second half from “21” to “12” and compare it with the first half “12”. Since they are equal, we know that $x$ is a palindrome.

Let’s see how to reverse the second half.

For the number $1221$, if we perform $1221 \bmod 10$, we will get the last digit $1$. To get the second last digit, we can first remove the last digit from $1221$ by dividing by $10$, $1221 / 10 = 122$, then get the remainder of the previous result divided by $10$, $122 \bmod 10 = 2$, to get the second last digit.

If we continue this process, we will get more reversed digits.

By continuously multiplying the last digit to the variable $y$, we can get the number in reverse order.

In the code implementation, we can repeatedly “take out” the last digit of $x$ and “add” it to the end of $y$, loop until $y \ge x$. If at this time $x = y$, or $x = y / 10$, then $x$ is a palindrome.

The time complexity is $O(\log_{10}(n))$, where $n$ is $x$. For each iteration, we will divide the input by $10$, so the time complexity is $O(\log_{10}(n))$. The space complexity is $O(1)$.

• class Solution {
public boolean isPalindrome(int x) {
if (x < 0 || (x > 0 && x % 10 == 0)) {
return false;
}
int y = 0;
for (; y < x; x /= 10) {
y = y * 10 + x % 10;
}
return x == y || x == y / 10;
}
}

• class Solution {
public:
bool isPalindrome(int x) {
if (x < 0 || (x && x % 10 == 0)) {
return false;
}
int y = 0;
for (; y < x; x /= 10) {
y = y * 10 + x % 10;
}
return x == y || x == y / 10;
}
};

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


• func isPalindrome(x int) bool {
if x < 0 || (x > 0 && x%10 == 0) {
return false
}
y := 0
for ; y < x; x /= 10 {
y = y*10 + x%10
}
return x == y || x == y/10
}

• function isPalindrome(x: number): boolean {
if (x < 0 || (x > 0 && x % 10 === 0)) {
return false;
}
let y = 0;
for (; y < x; x = ~~(x / 10)) {
y = y * 10 + (x % 10);
}
return x === y || x === ~~(y / 10);
}


• /**
* @param {number} x
* @return {boolean}
*/
var isPalindrome = function (x) {
if (x < 0 || (x > 0 && x % 10 === 0)) {
return false;
}
let y = 0;
for (; y < x; x = ~~(x / 10)) {
y = y * 10 + (x % 10);
}
return x === y || x === ~~(y / 10);
};


• impl Solution {
pub fn is_palindrome(mut x: i32) -> bool {
if x < 0 || (x % 10 == 0 && x != 0) {
return false;
}
let mut y = 0;
while x > y {
y *= 10;
y += x % 10;
x /= 10;
}
x == y || x == y / 10
}
}


• class Solution {
/**
* @param int $x * @return boolean */ function isPalindrome($x) {
$str = (string)$x;
$str_reverse = strrev($str);
return $str ===$str_reverse;
}
}