Skip to content

190. Reverse Bits

Description

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.

 

Example 1:

Input: n = 43261596

Output: 964176192

Explanation:

Integer Binary
43261596 00000010100101000001111010011100
964176192 00111001011110000010100101000000

Example 2:

Input: n = 2147483644

Output: 1073741822

Explanation:

Integer Binary
2147483644 01111111111111111111111111111100
1073741822 00111111111111111111111111111110

 

Constraints:

  • 0 <= n <= 231 - 2
  • n is even.

 

Follow up: If this function is called many times, how would you optimize it?

Solutions

Solution 1: Bit Manipulation

We can extract each bit of n from the least significant bit to the most significant bit, and then place it in the corresponding position of ans.

For example, for the \(i\)-th bit, we can use (n & 1) << (31 - i) to extract the \(i\)-th bit of n and place it on the \(31 - i\)-th bit of ans, then right shift n by one bit.

The time complexity is \(O(\log n)\), and the space complexity is \(O(1)\).

1
2
3
4
5
6
7
class Solution:
    def reverseBits(self, n: int) -> int:
        ans = 0
        for i in range(32):
            ans |= (n & 1) << (31 - i)
            n >>= 1
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
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 && n != 0; ++i) {
            ans |= (n & 1) << (31 - i);
            n >>>= 1;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t ans = 0;
        for (int i = 0; i < 32 && n; ++i) {
            ans |= (n & 1) << (31 - i);
            n >>= 1;
        }
        return ans;
    }
};

1
2
3
4
5
6
7
func reverseBits(n uint32) (ans uint32) {
    for i := 0; i < 32; i++ {
        ans |= (n & 1) << (31 - i)
        n >>= 1
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
impl Solution {
    pub fn reverse_bits(mut n: u32) -> u32 {
        let mut ans = 0;
        for i in 0..32 {
            ans |= (n & 1) << (31 - i);
            n >>= 1;
        }
        ans
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
/**
 * @param {number} n - a positive integer
 * @return {number} - a positive integer
 */
var reverseBits = function (n) {
    let ans = 0;
    for (let i = 0; i < 32 && n; ++i) {
        ans |= (n & 1) << (31 - i);
        n >>= 1;
    }
    return ans >>> 0;
};

Comments