Reverse an integer without using string operations.

Question

Randy Hammon · Accepted Answer

I give up on copy/pasting code. Keeps mangling it. The basic idea is a while statement with orig > 10.
Inside of loop add (mod 10 of orig) to reversed. multiply reversed by 10, divide orig  by 10. Last digit will be after exiting loop. Just add remaining orig to reversed. There is an extra check for a negative number. Really easy once you figure out the mod 10 idea.

Randy Hammon · Answer

yet again mangled.  while orig greater than or equal to 10.

Randy Hammon · Answer

That was fun.

Randy Hammon · Answer

Oops. answer would help. :)
#include 
#include 

using namespace std;

/*
 * 
 */
int main(int argc, char** argv) {
    int orig = -12345;
    int reversed;
    bool neg = false;
    cout  10) {
        reversed += orig % 10;
        reversed *= 10;
        orig /= 10;
    }
    reversed += orig;

    if (neg) {
        reversed *= -1;
    }
    cout << "reversed = " << reversed << "
";
    return 0;
}

Randy Hammon · Answer

Ugh. Code got mangled. Sure wish I could delete an errant comment.

#include 
#include 

using namespace std;

/*
 * 
 */
int main(int argc, char** argv) {
    int orig = 12345;
    int reversed;
    bool neg = false;
    cout  10) {
        reversed += orig % 10;
        reversed *= 10;
        orig /= 10;
    }
    reversed += orig;

    if (neg) {
        reversed *= -1;
    }
    cout << "reversed = " << reversed << "
";
    return 0;
}

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