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I assume by reverse, you mean the following:
123=>321
-123=>-321
+123=>+321

If the input is 'string', you might need some string manipulation.
Trim any extra spaces,commas etc.
Parse the leftmost char, if it's +/-, preserve it some stack.
Then try the following algo.
newnum=0;
while(num)
{
i=num%10;
if(newnum==0)
{
newnum=i;
}
else
{
newnum=newnum*10+i;
}
num=num/10;
}
return newnum;

can any one pls exp what is "reverse"

also, what is -ve and +ve?

hey wat if i take num=0321. Then it shud convert it to 1230.
i tried my code but it is not converting the number to 1230.
and same problem i found with ur code.

You are correct, any integer reversed twice should be equal to its original value. The reason you are experiencing all these problems is because you are reversing characters of a corresponding string representation of an integer, instead of reversing bits of that integer. If you look at binary representation of 2 (00000010) and truly reverse it to 01000000, you will arrive to decimal 64. Then, if you reverse 64 (01000000) back to 00000010, you will have your decimal 2 back.

why not simply convert the integer to string and reverse the string and then convert the string to integer. :0

int reverseDigits(int digit)
{
	int isNegative = 0;
	if(digit<0) {
		isNegative =1;
		digit*=-1;
	}
	if(digit==0) return 0;

	int reverse=0;
	while(digit){
		reverse*=10;
		reverse += digit%10;		
		digit/=10;		
	}	
	return isNegative==0? reverse:reverse*-1;
	
}

int reverse=0;
while(digit>0){

then it will work

How about this:

Let N = 2350.

while(N)
{
int x = N%10;
push(S, x); //push x into a stack S.
N /= 10;
k++ ; //keeps track of length of N
}

//now do pop operation
while(k--)
{
printf( "%d", pop(S));
}

+ve and -ve are abbreviations for “positive” and “negative”.