Reverse actual bits of the given number
Given a non-negative integer n. The problem is to reverse the bits of n and print the number obtained after reversing the bits. Note that the actual binary representation of the number is being considered for reversing the bits, no leading 0’s are being considered.
Examples :
Input : 11
Output : 13
(11)10 = (1011)2.
After reversing the bits we get:
(1101)2 = (13)10.
Input : 10
Output : 5
(10)10 = (1010)2.
After reversing the bits we get:
(0101)2 = (101)2
= (5)10.
In this approach, one by one bits in binary representation of n are being obtained with the help of bitwise right shift operation and they are being accumulated in rev with the help of bitwise left shift operation.
Algorithm:
C++
// C++ implementation to reverse bits of a number #include <bits/stdc++.h> using namespace std; // function to reverse bits of a number unsigned int reverseBits(unsigned int n) { unsigned int rev = 0; // traversing bits of 'n' from the right while (n > 0) { // bitwise left shift // 'rev' by 1 rev <<= 1; // if current bit is '1' if (n & 1 == 1) rev ^= 1; // bitwise right shift // 'n' by 1 n >>= 1; } // required number return rev; } // Driver program to test above int main() { unsigned int n = 11; cout << reverseBits(n); return 0; } |
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Java
// Java implementation to // reverse bits of a number class GFG { // function to reverse bits of a number public static int reverseBits(int n) { int rev = 0; // traversing bits of 'n' // from the right while (n > 0) { // bitwise left shift // 'rev' by 1 rev <<= 1; // if current bit is '1' if ((int)(n & 1) == 1) rev ^= 1; // bitwise right shift //'n' by 1 n >>= 1; } // required number return rev; } // Driver code public static void main(String[] args) { int n = 11; System.out.println(reverseBits(n)); } } // This code is contributed // by prerna saini. |
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Python 3
# Python 3 implementation to # reverse bits of a number # function to reverse # bits of a number def reverseBits(n) : rev = 0 # traversing bits of 'n' from the right while (n > 0) : # bitwise left shift 'rev' by 1 rev = rev << 1 # if current bit is '1' if (n & 1 == 1) : rev = rev ^ 1 # bitwise right shift 'n' by 1 n = n >> 1 # required number return rev # Driver code n = 11print(reverseBits(n)) # This code is contributed # by Nikita Tiwari. |
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C#
// C# implementation to // reverse bits of a number using System; class GFG { // function to reverse bits of a number public static int reverseBits(int n) { int rev = 0; // traversing bits of 'n' // from the right while (n > 0) { // bitwise left shift // 'rev' by 1 rev <<= 1; // if current bit is '1' if ((int)(n & 1) == 1) rev ^= 1; // bitwise right shift //'n' by 1 n >>= 1; } // required number return rev; } // Driver code public static void Main() { int n = 11; Console.WriteLine(reverseBits(n)); } } // This code is contributed // by vt_m. |
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PHP
<?php // PHP implementation to reverse // bits of a number // function to reverse // bits of a number function reverseBits($n) { $rev = 0; // traversing bits of 'n' // from the right while ($n > 0) { // bitwise left shift // 'rev' by 1 $rev <<= 1; // if current bit is '1' if ($n & 1 == 1) $rev ^= 1; // bitwise right shift // 'n' by 1 $n >>= 1; } // required number return $rev; } // Driver code $n = 11; echo reverseBits($n); // This code is contributed by mits ?> |
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Output :
13
Time Complexity: O(num), where num is the number of bits in the binary representation of n.
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Improved By : Mithun Kumar
