检查数字的实际二进制表示形式是否是回文

给定一个非负整数n 。问题是要检查n的二进制表示形式是否是回文。请注意，正考虑使用数字的实际二进制表示进行回文检查，而不考虑前导0。
例子 ：

Input : 9
Output : Yes
(9)10 = (1001)2

Input : 10
Output : No
(10)10 = (1010)2

方法：以下是步骤：

1. 获取通过反转n的二进制表示形式的位获得的数字。请参阅这篇文章。让它成为rev 。
2. 如果n == rev，则打印“是”，否则打印“否”。

C++

// C++ implementation to check whether binary
// representation of a number is palindrome or not
#include 
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

 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

 
// 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");
 

输出 ：

Yes

时间复杂度：O(num)，其中num是n的二进制表示形式中的位数。