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 $O(1)$ .

Python Code

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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)

Java Code

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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;
    }
}

C++ Code

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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;
    }
};

Go Code

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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
}

TypeScript Code

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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);
}

Rust Code

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impl Solution {
    pub fn is_palindrome(x: i32) -> bool {
        if x < 0 {
            return false;
        }
        let s = x.to_string();
        let bs = s.as_bytes();
        let n = bs.len();
        let mut l = 0;
        let mut r = n - 1;
        while l < r {
            if bs[l] != bs[r] {
                return false;
            }
            l += 1;
            r -= 1;
        }
        true
    }
}

Rust Code

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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
    }
}

JavaScript Code

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/**
 * @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);
};

PHP Code

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class Solution {
    /**
     * @param int $x
     * @return boolean
     */

    function isPalindrome($x) {
        $str = (string) $x;
        $str_reverse = strrev($str);
        return $str === $str_reverse;
    }
}

9. Palindrome Number#

Description#

Solutions#

Solution 1: Reverse Half of the Number#

9. Palindrome Number

Description

Solutions

Solution 1: Reverse Half of the Number