Dado un entero no negativo n . El problema es comprobar si la representación binaria de n es palíndromo o no. Tenga en cuenta que la representación binaria real del número se está considerando para la verificación del palíndromo, no se están considerando los 0 iniciales.
Ejemplos:
Input : 9 Output : Yes (9)10 = (1001)2 Input : 10 Output : No (10)10 = (1010)2
Enfoque: Los siguientes son los pasos:
- Obtenga el número obtenido al invertir los bits en la representación binaria de n . Consulte esta publicación. Que sea rev .
- Si n == rev, imprima «Sí», de lo contrario, «No».
C++
// C++ implementation to check whether binary
// representation of a number is palindrome or not
#include <bits/stdc++.h>
using namespace std;
// function to reverse bits of a number
unsigned int reverseBits(unsigned int n)
{
unsigned int rev = 0;
// traversing bits of 'n' from the right
while (n > 0) {
// bitwise left shift 'rev' by 1
rev <<= 1;
// if current bit is '1'
if (n & 1 == 1)
rev ^= 1;
// bitwise right shift 'n' by 1
n >>= 1;
}
// required number
return rev;
}
// function to check whether binary representation
// of a number is palindrome or not
bool isPalindrome(unsigned int n)
{
// get the number by reversing bits in the
// binary representation of 'n'
unsigned int rev = reverseBits(n);
return (n == rev);
}
// Driver program to test above
int main()
{
unsigned int n = 9;
if (isPalindrome(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java code to check whether
// binary representation of a
// number is palindrome or not
import java.util.*;
import java.lang.*;
public class GfG
{
// function to reverse bits of a number
public static long reverseBits(long n)
{
long rev = 0;
// traversing bits of 'n'
// from the right
while (n > 0)
{
// bitwise left shift 'rev' by 1
rev <<= 1;
// if current bit is '1'
if ((n & 1) == 1)
rev ^= 1;
// bitwise right shift 'n' by 1
n >>= 1;
}
// required number
return rev;
}
// function to check a number
// is palindrome or not
public static boolean isPalindrome(long n)
{
// get the number by reversing
// bits in the binary
// representation of 'n'
long rev = reverseBits(n);
return (n == rev);
}
// Driver function
public static void main(String argc[])
{
long n = 9;
if (isPalindrome(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Sagar Shukla
Python3
# Python 3 implementation to check
# whether binary representation of
# a number is palindrome or not
# function to reverse bits of a number
def reverseBits(n) :
rev = 0
# traversing bits of 'n' from the right
while (n > 0) :
# bitwise left shift 'rev' by 1
rev = rev << 1
# if current bit is '1'
if (n & 1 == 1) :
rev = rev ^ 1
# bitwise right shift 'n' by 1
n = n >> 1
# required number
return rev
# function to check whether binary
# representation of a number is
# palindrome or not
def isPalindrome(n) :
# get the number by reversing
# bits in the binary
# representation of 'n'
rev = reverseBits(n)
return (n == rev)
# Driver program to test above
n = 9
if (isPalindrome(n)) :
print("Yes")
else :
print("No")
# This code is contributed by Nikita Tiwari.
C#
// C# code to check whether
// binary representation of a
// number is palindrome or not
using System;
public class GfG
{
// function to reverse bits of a number
public static long reverseBits(long n)
{
long rev = 0;
// traversing bits of 'n'
// from the right
while (n > 0)
{
// bitwise left shift 'rev' by 1
rev <<= 1;
// if current bit is '1'
if ((n & 1) == 1)
rev ^= 1;
// bitwise right shift 'n' by 1
n >>= 1;
}
// required number
return rev;
}
// function to check a number
// is palindrome or not
public static bool isPalindrome(long n)
{
// get the number by reversing
// bits in the binary
// representation of 'n'
long rev = reverseBits(n);
return (n == rev);
}
// Driver function
public static void Main()
{
long n = 9;
if (isPalindrome(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m
PHP
<?php
// PHP implementation to check
// whether binary representation
// of a number is palindrome or not
// function to reverse bits of a number
function reverseBits($n)
{
$rev = 0;
// traversing bits of 'n'
// from the right
while ($n > 0)
{
// bitwise left shift 'rev' by 1
$rev <<= 1;
// if current bit is '1'
if ($n & 1 == 1)
$rev ^= 1;
// bitwise right shift 'n' by 1
$n >>= 1;
}
// required number
return $rev;
}
// function to check whether
// binary representation of a
// number is palindrome or not
function isPalindrome($n)
{
// get the number by reversing
// bits in the binary representation
// of 'n'
$rev = reverseBits($n);
return ($n == $rev);
}
// Driver code
$n = 9;
if (isPalindrome($n))
echo "Yes";
else
echo "No";
return 0;
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript program to check whether
// binary representation of a
// number is palindrome or not
// function to reverse bits of a number
function reverseBits(n)
{
let rev = 0;
// traversing bits of 'n'
// from the right
while (n > 0)
{
// bitwise left shift 'rev' by 1
rev <<= 1;
// if current bit is '1'
if ((n & 1) == 1)
rev ^= 1;
// bitwise right shift 'n' by 1
n >>= 1;
}
// required number
return rev;
}
// function to check a number
// is palindrome or not
function isPalindrome(n)
{
// get the number by reversing
// bits in the binary
// representation of 'n'
let rev = reverseBits(n);
return (n == rev);
}
// Driver code
let n = 9;
if (isPalindrome(n))
document.write("Yes");
else
document.write("No");
</script>
Producción :
Yes
Complejidad de tiempo: O(num), donde num es el número de bits en la representación binaria de n .
Espacio Auxiliar: O(1).
Publicación traducida automáticamente
Artículo escrito por ayushjauhari14 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA