Welcome to Subscribe On Youtube

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 $1221mod10$, 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 $10$, $122mod10=2$, 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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript
• Php
• C#

• 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;
      }
  }
  
  

• public class Solution {
      public bool 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;
      }
  }
  

All Problems

All Solutions