Reversing a Number using bitwise shift

Question

I am trying to find a way to reverse a number without

Converting it to a string to find the length
Reversing the string and parsing it back
Running a separate loop to compute the Length

i am currently doing it this way

 public static int getReverse(int num){ int revnum =0; for( int i = Integer.toString(num).length() - 1 ; num>0 ; i-- ){ revnum += num % 10 * Math.pow( 10 , i ); num /= 10; } return revnum; }

But I would Like to implement the above 3 conditions.

I am looking for a way , possibly using the bit wise shift operators or some other kind of bitwise operation.

Is it possible ? If so how ?

PS : If 1234 is given as input it should return 4321. I will only be reversing Integers and Longs

How would you reverse 10000? Only 1? You will loose the "zeros".. — dacwe
– dacwe, Commented Sep 19, 2011 at 14:56
You are unlikely to find a good way to reverse the decimal representation with bitwise operations. — Oliver Charlesworth
– Oliver Charlesworth, Commented Sep 19, 2011 at 15:00
"using bitwise shift" There is absolutely no bitwise shifting going on here. — Andreas
– Andreas, Commented May 1, 2021 at 21:28

NPE · Accepted Answer · 2011-09-19 15:04:14Z

7

How about:

int revnum = 0; while (num != 0) { revnum = revnum * 10 + (num % 10); num /= 10; } return revnum;

The code expects a non-negative input.

This may or may not matter to you, but it's worth noting that getReverse(getReverse(x)) does not necessarily equal x as it won't preserve trailing zeroes.

edited Sep 19, 2011 at 15:04

answered Sep 19, 2011 at 14:57

NPE

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3 Comments

Gautam Over a year ago

Its ok If it doesn't reverse negative numbers , I'll just convert it into positive and reverse it and then convert it back to negative

Gautam Over a year ago

I guess getReverse(getReverse(x)) won't be much of a problem now , but if it does I might have to rethink what I am using this function for. I think I will be getting the same problem even with the previous implementation

NPE Over a year ago

@gautham5678: That's correct: this is a feature of the function's signature (int->int) rather than of this implementation.

dreamlax · Accepted Answer · 2013-03-26 21:36:19Z

How about this? It handles negative numbers as well.

public int getReverse(int num){ int rst=0; int sign; sign=num>0?1:-1; num*=sign; while(num>0){ int lastNum = num%10; rst=rst*10+lastNum num=num/10; } return rst*sign; }

Koushlendra · Accepted Answer · 2021-05-01 21:45:05Z

Integer.MAX_VALUE or Integer.MIN_VALUE is not at all considered in any of the solutions.

For eg: if the input is -2147483647 or 2147483647 the o/p will be 1126087180 and -1126087180 respectively.

Please try the below solutions where both the conditions are handled, so if at any point of time the number is going beyond the boundary conditions i.e., INPUT_VALUE > Integer.MAX_VALUE or INPUT_VALUE < Integer.MIN_VALUE it would return 0

class ReversingIntegerNumber { public int reverse(int originalNum) { boolean originalIsNegative = originalNum > 0 ? false : true; int reverseNum = 0; int modValue; originalNum = Math.abs(originalNum); while(originalNum != 0) { modValue = originalNum % 10; if(reverseNum > (Integer.MAX_VALUE - modValue)/10) { return 0; } reverseNum = (reverseNum * 10) + modValue; originalNum /= 10; } return originalIsNegative ? -1 * reverseNum : reverseNum; } }

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