Reversing a Number using bitwise shift
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 currentl开发者_如何学Cy 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 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.
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;
}
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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