Is this number a palindrome?

Question

Input

A single integer in your language's preferred format. This format is non-negotiable (no input as strings, arrays of digits, etc.), but standard I/O formats do apply if otherwise it would be impossible in your language.

Output

A truthy or falsely value, depending on whether the given number is palindromic (same forwards and backwards)

Rules

• Numbers will always be positive, at least two digits long.
• This is fastest-algorithm, so lowest O(n) time complexity wins.

Test cases

• 123454321 - true
• 296296 - false
• 296692 - true
• 5499456 - false
• 0 - false/error/any value

Answer 1

APL (Dyalog Unicode), O(1)

⊃∘'111111111110… …10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000'

Anonymous tacit prefix function.

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Simply picks the result from a pre-compiled list.

Answer 2

Java, complexity: \$O(\lfloor\log_{10}(n) + 1\rfloor)\$

boolean f(int n){
  int reverse = 0;
  //int debugIterations = 0;
  for(int palindrome = n; palindrome != 0; palindrome /= 10){
    //debugIterations++;
    int lastDigit = palindrome % 10;
    reverse = reverse * 10 + lastDigit;
  }
  //System.out.println("Debug iterations: "+debugIterations);
  return n == reverse;
}

Loops once for every digit in the input-number. Pretty straight-forward implementation tbh..

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Answer 3

Java, O(1)

boolean f(int n){
  if (n <= 0 || n >= 16777216) throw new Error(n+" is not a valid input!");
  int reverse = n/10000000%10*1 + n/1000000%10*10 + n/100000%10*100 + n/10000%10*1000 + n/1000%10*10000 + n/100%10*100000 + n/10%10*1000000 + n/1%10*10000000;
  int len = (int) Math.log10(n);
  reverse/= Math.pow(10, 7-len);
  return reverse == n;
}

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Kevin Cruijssen's answer, unrolled for 24-bit numbers.

Answer 4

APL+WIN, O(1)

Thanks to Adam for suggested changes to my original code and his explanation - see the comments section.

Prompts for input of integer.

 17=+/(⌽17↑n)=¯17↑n←⍕⎕

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