notes/OJ notes/pages/Leetcode Reverse-Bits.md

Leetcode Reverse-Bits

2022-07-22 15:15

Algorithms:

#algorithm #bit_manipulation

Data structures:

#DS #bitset

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem
• Solution and detailed explanation
• Shifting solution an explanation

---

Problem

Reverse bits of a given 32 bits unsigned integer.

Examples

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

Thoughts

[!summary] This is a #bit_manipulation problem.

There are two methods:

• swapping bits inside the bitset
• shifting

Method 1: Swapping

This is the most simple and intuitive one, all you have to do is swap the first and last numbers, and continue like that.

Method 2: Shifting

The bitset has operations like this << >>, which can be used to shift, and reverse the array.

The operation is already documented in the link above: Leetcode Reverse-Bits#Links

Solution

Method 1:

class Solution {
public:
  uint32_t reverseBits(uint32_t n) {
    // swap bits inside the function
    bitset<32> nb(n);
    bool tmp;
    int l = 0, r = 31;

    while (l < r) {
      tmp = nb[l];
      nb[l] = nb[r];
      nb[r] = tmp;
      l++;
      r--;
    }

    return nb.to_ulong();
  }
};

Shifting:

==#TODO solve it using shifting==