Interview Question for Software Development Engineer II at Amazon.com:

This was the first question I was asked and is considered a warm up.

public static boolean isPalindrome(int someInt) {
        final int WIDTH = 8;
        boolean isPalindrome = true;
        for (int i = 0; i < WIDTH && isPalindrome == true; i++) {

            int maskLower = (int) Math.pow(2, i);
            int maskUpper = (int) Math.pow(2, WIDTH - (i+1));

            boolean bitLowerOn = ((maskLower & someInt) == maskLower) ? true : false;
            boolean bitUpperOn = ((maskUpper & someInt) == maskUpper) ? true : false;

            isPalindrome = bitLowerOn && bitUpperOn && isPalindrome || !bitLowerOn && !bitUpperOn;
        }
        return isPalindrome;
    }

anon.. would this work for a number like 17 (10001)?

bool checkPalindrome(unsigned int n)
{
    int m = n, k =0;
    bool ret = false;
    while(m!=0)
    {
        int i = 1;
        i = i & m;
        k = k << 1;
        k = k | i;
        m = m >> 1;
    }
    if((k^n)==0)
    {
        cout<<"Palindrome"<<endl;
        ret = true;
    }
    else
    {
        cout<<"Not Palindrome"<<endl;
    }

    return ret;
}

I have a simple solution. Reverse the bits one by one and test equality of the two resulting numbers. (Matlab code)

function [r] = isPalindrome(a)

n = a;
m = 0;

while(n>0)
    m = bitshift(m, 1);
    m = m + mod(n, 2);
    n = bitshift(n, -1);
end

r = (m==a);

end

public static boolean isPalindrome(int n) {
        int nb = 0, nl=n;
        while(nl>0) {
            nb=(nb<<1) + (nl&1);
            nl=nl>>1;
        }
        return nb==n;
    }