Issue
I have been trying this for some time now but could not get it to work. I am trying to have a method to reverse an integer without the use of strings or arrays. For example, 123 should reverse to 321 in integer form.
My first attempt:
/** reverses digits of integer using recursion */
public int RevDigs(int input)
{
int reverse = 0;
if(input == 0)
{
return reverse;
}
int tempRev = RevDigs(input/10);
if(tempRev >= 10)
reverse = input%10 * (int)Math.pow(tempRev/10, 2) + tempRev;
if(tempRev <10 && tempRev >0)
reverse = input%10*10 + tempRev;
if(tempRev == 0)
reverse = input%10;
return reverse;
}//======================
I also tried to use this, but it seems to mess up middle digits:
/** reverses digits of integer using recursion */
public int RevDigs(int input)
{
int reverse = 0;
if(input == 0)
{
return reverse;
}
if(RevDigs(input/10) == 0)
reverse = input % 10;
else
{
if(RevDigs(input/10) < 10)
reverse = (input % 10) *10 + RevDigs(input/10);
else
reverse = (input % 10)* 10 * (RevDigs(input/10)/10 + 1) + RevDigs(input/10);
}
return reverse;
}
I have tried looking at some examples on the site, however I could not get them to work properly. To further clarify, I cannot use a String, or array for this project, and must use recursion. Could someone please help me to fix the problem. Thank you.
Solution
What about:
public int RevDigs(int input) {
if(input < 10) {
return input;
}
else {
return (input % 10) * (int) Math.pow(10, (int) Math.log10(input)) + RevDigs(input/10);
/* here we:
- take last digit of input
- multiply by an adequate power of ten
(to set this digit in a "right place" of result)
- add input without last digit, reversed
*/
}
}
This assumes input >= 0
, of course.
Answered By - Michał Rybak
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