Well there are multiple ways to reverse the bits of the given number in Java.
First, Java language has inbuild bitwise complement operator(~). So (~number) reverses the bits of number.
Second, one can use the Integer.reverse(number)
Third, if this is a part of test or you just want to play with bits, you can refer the code below.
/*
The logic uses moving bitmask from right to left:
1. Get the bit of given number, by binary and(&) with bitmask
2. XOR(^) with the bitmask, so here we reverse the bit.
3. OR(|) this reversed bit with the result(result has all Zero initially)
This logic is repeated for each 32 bits by moving the mask from right to left,
one bit at a time using (<<) left shift operator on bitmask.
*/
public class ReverseBits {
public static int reverseBits(int input) {
print("Input", input);
int bitmask = 1;
int result = 0;
do {
//print("Bitmask", bitmask);
result = result | (bitmask ^ (input & bitmask)) ;
//print("Result", result);
bitmask = bitmask << 1;
} while (bitmask != 0);
print("Reverse", result);
return result;
}
public static void print(String label, int input) {
System.out.println(label +"\t:"+Integer.toBinaryString(input));
}
public static void main(String[] args) {
reverseBits(Integer.MIN_VALUE);
reverseBits(Integer.MAX_VALUE);
reverseBits(reverseBits(170));
}
}
Output:
Input :10000000000000000000000000000000
Reverse :1111111111111111111111111111111
Input :1111111111111111111111111111111
Reverse :10000000000000000000000000000000
Input :10101010
Reverse :11111111111111111111111101010101
Input :11111111111111111111111101010101
Reverse :10101010