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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;
        }
    }
    
    

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