Write a palindrome-checking function

Question

Tristan · Accepted Answer

This has a N/2 complexity :

function isPalyndrome($str) {
	$array = str_split($str);
	$size = count($array);
	$pivot = floor($size / 2);

	for($i = 0; $i < $pivot; $i++) {
		if($array[$i] != $array[$size - $i - 1]) {
			return false;
		}
	}

	return true;
}

Bikash · Answer

#include 
#include 

using namespace std;

bool IsPalindrome(const string& input) {
  for (int i = 0; i < input.size()/2; ++i) {
 if (input[i] != input[input.size()-i-1]) {
   return false;
 }
  }
  return true;
}

int main() {
  const string str("madam");
  const string str1("madama");
  cout << IsPalindrome(str) << endl;
  cout << IsPalindrome(str1) << endl;
  
  return 0;
}

Utilisateur anonyme · Answer

My solution walked char pointers back from the end and forward from the beginning of the string, returning true if the crossed over and false if the two chars pointed to didn't match.

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