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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

• Java
• C++
• Python
• Go
• Javascript
• RenderScript

• 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;
      }
  };
  

• class Solution:
      def reverseBits(self, n: int) -> int:
          res = 0
          for i in range(32):
              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
  }
  

• /**
   * @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;
      }
      return res >>> 0;
  };
  
  

• 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
      }
  }
  
  

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