Dado un número ‘n’, nuestro objetivo es averiguar si es palíndromo o no sin usar
ningún espacio adicional. No podemos hacer una nueva copia del número.
Ejemplos:
Input : 2332 Output : Yes it is Palindrome. Explanation: original number = 2332 reversed number = 2332 Both are same hence the number is palindrome. Input :1111 Output :Yes it is Palindrome. Input : 1234 Output : No not Palindrome.
Una solución recursiva se discute en la publicación a continuación.
Comprobar si un número es palíndromo
En esta publicación se analiza una solución diferente.
1) Podemos comparar el primer dígito y el último dígito, luego repetimos el proceso.
2) Para el primer dígito, necesitamos el orden del número. Digamos, 12321. Dividir esto por 10000 nos daría el 1 principal. El 1 final se puede recuperar tomando el mod con 10.
3) Ahora, para reducir esto a 232.
(12321 % 10000)/10 = (2321)/10 = 232
4) Y ahora, el 10000 tendría que reducirse en un factor de 100.
Aquí está la implementación del algoritmo anterior:
C++
// C++ program to find number is palindrome
// or not without using any extra space
#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(int);
bool isPalindrome(int n)
{
// Find the appropriate divisor
// to extract the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
int main()
{
isPalindrome(1001) ? cout << "Yes, it is Palindrome" :
cout << "No, not Palindrome";
return 0;
}
Java
// Java program to find number is palindrome
// or not without using any extra space
public class GFG
{
static boolean isPalindrome(int n)
{
// Find the appropriate divisor
// to extract the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
public static void main(String args[])
{
if(isPalindrome(1001))
System.out.println("Yes, it is Palindrome");
else
System.out.println("No, not Palindrome");
}
}
// This code is contributed by Sumit Ghosh
Python3
# Python program to find number
# is palindrome or not without
# using any extra space
# Function to check if given number
# is palindrome or not without
# using the extra space
def isPalindrome(n):
# Find the appropriate divisor
# to extract the leading digit
divisor = 1
while (n / divisor >= 10):
divisor *= 10
while (n != 0):
leading = n // divisor
trailing = n % 10
# If first and last digit
# not same return false
if (leading != trailing):
return False
# Removing the leading and
# trailing digit from number
n = (n % divisor)//10
# Reducing divisor by a factor
# of 2 as 2 digits are dropped
divisor = divisor/100
return True
# Driver code
if(isPalindrome(1001)):
print('Yes, it is palindrome')
else:
print('No, not palindrome')
# This code is contributed by Danish Raza
C#
// C# program to find number
// is palindrome or not without
// using any extra space
using System;
class GFG
{
static bool isPalindrome(int n)
{
// Find the appropriate
// divisor to extract
// the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and
// trailing digit from number
n = (n % divisor) / 10;
// Reducing divisor by
// a factor of 2 as 2
// digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
static public void Main ()
{
if(isPalindrome(1001))
Console.WriteLine("Yes, it " +
"is Palindrome");
else
Console.WriteLine("No, not " +
"Palindrome");
}
}
// This code is contributed by m_kit
PHP
<?php
// PHP program to find
// number is palindrome
// or not without using
// any extra space
function isPalindrome($n)
{
// Find the appropriate divisor
// to extract the leading digit
$divisor = 1;
while ($n / $divisor >= 10)
$divisor *= 10;
while ($n != 0)
{
$leading = floor($n / $divisor);
$trailing = $n % 10;
// If first and last digit
// not same return false
if ($leading != $trailing)
return false;
// Removing the leading and
// trailing digit from number
$n = ($n % $divisor) / 10;
// Reducing divisor by a
// factor of 2 as 2 digits
// are dropped
$divisor = $divisor / 100;
}
return true;
}
// Driver code
if(isPalindrome(1001) == true)
echo "Yes, it is Palindrome" ;
else
echo "No, not Palindrome";
// This code is contributed by ajit
?>
Javascript
<script>
// javascript program to find number is palindrome
// or not without using any extra space
function isPalindrome(n) {
// Find the appropriate divisor
// to extract the leading digit
var divisor = 1;
while (parseInt(n / divisor) >= 10)
divisor *= 10;
while (n != 0) {
var leading = parseInt(n / divisor);
var trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = parseInt((n % divisor) / 10);
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Driver code
if (isPalindrome(1001))
document.write("Yes, it is Palindrome");
else
document.write("No, not Palindrome");
// This code is contributed by todaysgaurav
</script>
Producción:
Yes, it is Palindrome
Este artículo es una contribución de Abhijit Shankhdhar . Si le gusta GeeksforGeeks y le gustaría contribuir, también puede escribir un artículo usando contribuya.geeksforgeeks.org o envíe su artículo por correo a contribuya@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA