Question

190. Reverse Bits

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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

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

public class Solution {
  // you need treat n as an unsigned value
  public int reverseBits(int n) {
    int res = 0;
    for (int i = 0; i < 32 && n != 0; ++i) {
      res |= ((n & 1) << (31 - i));
      n >>>= 1;
    }
    return res;
  }
}

class Solution {
public:
  uint32_t reverseBits(uint32_t n) {
    uint32_t res = 0;
    for (int i = 0; i < 32; ++i) {
      res |= ((n & 1) << (31 - i));
      n >>= 1;
    }
    return res;
  }
};

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

impl Solution {
  pub fn reverse_bits(mut x: u32) -> u32 {
    let mut res = 0;
    for _ in 0..32 {
      res = (res << 1) | (x & 1);
      x >>= 1;
    }
    res
  }
}

/**
 * @param {number} n - a positive integer
 * @return {number} - a positive integer
 */
var reverseBits = function (n) {
  let res = 0;
  for (let i = 0; i < 32 && n > 0; ++i) {
    res |= (n & 1) << (31 - i);
    n >>>= 1;
  }
  // 无符号右移
  return res >>> 0;
};