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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. 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

 

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 $O(1)$.

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

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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;
    }
}

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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;
    }
};

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

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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
    }
}

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/**
 * @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;
};

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