Leetcode

Reverse Bits

May 26, 2022 by aAshish

  • Time:O(32) = O(1)
  • Space:O(1)

C++

 
class Solution {
 public:
  uint32_t reverseBits(uint32_t n) {
    uint32_t ans = 0;

    for (int i = 0; i < 32; ++i)
      if (n >> i & 1)
        ans |= 1 << 31 - i;

    return ans;
  }
};

JAVA

 
public class Solution {
  // you need treat n as an unsigned value
  public int reverseBits(int n) {
    int ans = 0;

    for (int i = 0; i < 32; ++i)
      if ((n >> i & 1) == 1)
        ans |= 1 << 31 - i;

    return ans;
  }
}

Python

 
class Solution:
  def reverseBits(self, n: int) -> int:
    ans = 0

    for i in range(32):
      if n >> i & 1:
        ans |= 1 << 31 - i

    return ans