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@Anonymous: it doesn't matter. Here is my solution.

boolean isPalindrome(int x)
{
    int y=0,z=x;
    while(z)
    {
        y = (y<<1)|(z&1);
        z=z>>1;
    }
    return (y==x)?true:false;
}

Do we need to consider all bits in sizeof(int)??
Or only the number of digits required to represent that integer in binary notaion.
e.g. lets say sizeof int is 4 byte=32 bits
so 5 = 00000000000000000000000000000101
or we need to consider only 101

Not a efficient program, but works below code:
int n=0, num = 9;
int val = num;
while(num)
{
n = n << 1;
n |= (1&num);
num = num >> 1;
}
if(n == val)
cout << "Palindrome" << endl;
else
cout << "NOT Palindrome" << endl;

int ispalindrome(int x)
{
        int n=x,y=0;
        while(n)
        {
                y=(y<<1) | (n&1);
                n = n>>1;
        }
        return x==y;
}
 
int main()
{
        int a=3;
        printf(ispalindrome(a)==1?"Palindrome.":"Not a palindrome.");
}

HI Guys You Can No One Has Given The Algorithm Here , so if anyone interested to analyze that ,i coded & explained it here some time back , you can have a look , comment if anything wrong
shashank7s dot blogspot dot com/2011/05/wap-to-check-that-binary-representation dot html

Shashank

Here is an optimization for the case when the palindrome has an even number of bits.

boolean isPalindrome(int x)
{
   int z = x;
   int y = 0;
   while ((x != y) && (x!=0))
   {
      y = (y << 1) | (x & 1);
      x = x >> 1;
   }
   return x == y || z == y;
}

int palindrome(int x){
     int i ;
     for(i = 0; i < 32; i++){
         if( ( (x >> 31 - i) ^ ( x & (1 << i)) ) != 0)
             return 0;
     }
     return 1;
 }

#define TRUE 1
#define FALSE 0

int is_palindrome(int x)
{
    int reverse = 0;
    int result;
    int y = abs(x);

    while (y != 0) {
        reverse = reverse << 1;
        if ((y & 1))
            reverse = reverse | 1;

        y = y >> 1;
    }

    result = reverse & abs(x);
    if (x < 0) result = 0 - result;

    return (x == result) ? TRUE : FALSE;
}

#Python
{
def palindrome(a):
b=a[::-1]
if (b==a): return True
else: return False
}
}

#define isOne(num,pos)  (( num & ( 1 << pos ) ) ? TRUE : FALSE )
int isPalindrome( int num ) {
   for( pos = 15 ; pos != -1 ; pos-- )
      res = res + ( ( isOne(num,pos) ) ? ( 1 << (15 - pos) ) : 0 );
   return ( num == res );
}

This is another version......

#include<stdio.h>
void main()
{
unsigned int num = 12345;
unsigned char  rmb, lmb, ctr;
for(ctr=1; ctr<(sizeof(num) * 8)/2; ctr++) {
rmb = (num & (1<<((sizeof(num) * 8) - ctr ))) == 0 ? 0 : 1;
lmb = (num & (1<< (ctr-1))) > 0 ? 1 :0;
if(rmb!=lmb) break;
}
if(rmb==lmb) printf("Pali");
else printf("! Pali");
}

How about this logic

boolean is_palindrome(int x)
{
int comp=~x;
if( x&comp==0) return true ;
else false;
}