Algorithm


Problem Name: 190. Reverse Bits

Problem Link: https://leetcode.com/problems/reverse-bits/
 

Reverse bits of a given 32 bits unsigned integer.

Note:

  • Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
  • In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

 

Example 1:

Input: n = 00000010100101000001111010011100
Output:    964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

 

Constraints:

  • The input must be a binary string of length 32

Code Examples

#1 Code Example with C Programming

Code - C Programming


uint32_t reverseBits(uint32_t n) {
    /*
    uint32_t k, i;
    k = 0;
    for (i = 0; i  <  32; i ++) {
        k = (k << 1) | (n & 1);
        n = n >> 1;
    }
    return k;
    */
    n = ((n & 0xffff0000) >> 16) | ((n & 0x0000ffff) << 16);
    n = ((n & 0xff00ff00) >> 8)  | ((n & 0x00ff00ff) << 8);
    n = ((n & 0xf0f0f0f0) >> 4)  | ((n & 0x0f0f0f0f) << 4);
    n = ((n & 0xcccccccc) >> 2)  | ((n & 0x33333333) << 2);
    n = ((n & 0xaaaaaaaa) >> 1)  | ((n & 0x55555555) << 1);
    
    return n;
}
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Input

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n = 00000010100101000001111010011100

Output

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964176192 (00111001011110000010100101000000)

#2 Code Example with Java Programming

Code - Java Programming


public class Solution {
    // you need treat n as an unsigned value
  public int reverseBits(int n) {
    if (n == 0) {
      return 0;
    }
    int result = 0;
    for (int i = 0; i  <  32; i++) {
      result <<= 1;
      if ((n & 1) == 1) {
        result++;
      }
      n >>= 1;
    }
    return result;
  }
}
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Input

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n = 00000010100101000001111010011100

Output

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964176192 (00111001011110000010100101000000)

#3 Code Example with Javascript Programming

Code - Javascript Programming


const reverseBits = function(n) {
  let r = 0;
  for (let i = 0; i  <  32; i++) {
    if (n & 1) {
      r = r | 1;
    } 
    if (i !== 31) {
       r = r << 1;
       n = n >> 1;
    }
  }
  return r >>> 0;
};
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Input

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n = 11111111111111111111111111111101

Output

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3221225471 (10111111111111111111111111111111)

#4 Code Example with C# Programming

Code - C# Programming


namespace LeetCode
{
    public class _0190_ReverseBits
    {
        public uint reverseBits(uint n)
        {
            n = (n >> 16) | (n << 16);
            n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
            n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
            n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
            n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
            return n;
        }
    }
}
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Input

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n = 11111111111111111111111111111101

Output

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+
cmd
3221225471 (10111111111111111111111111111111)
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