Question

Formatted question description: https://leetcode.ca/all/190.html

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?

Algorithm

Take out the number to be flipped, from right to left. If it is 1, move the result to the left by one bit and add 1; if it is 0, move the result to the left by one bit, Then move n to the right.

Code

public class Reverse_Bits {

    public static void main(String[] args) {
        Reverse_Bits out = new Reverse_Bits();
        Solution s = out.new Solution();

        s.reverseBits(new Integer("11111111111111111111111111111101"));
    }

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

            int result = 0;

            for (int i = 0; i < 32; i++) {
                result += n & 1;

                // @note: >> 和 >>> 的区别，>>>左边补0，>>左边不补0
                n >>= 1;
                if (i < 31) { //@note: not <=
                    result <<= 1; // final bit, no left shift
                }
            }

            return result;
        }
    }



    public class Solution_overflow {
        // you need treat n as an unsigned value
        public int reverseBits(int n) {

            if (n < 0) {
                return -1;
            }

            String original = Integer.toBinaryString(n);
            String reversed = new StringBuilder(original).reverse().toString();

            // long cannot handle unsigned int
            long result = Long.parseLong(reversed);

            return (int)result;
        }
    }
}

############

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;
    }
}

// OJ: https://leetcode.com/problems/reverse-bits/
// Time: O(1)
// Space: O(1)
class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        int ans = 0;
        for (int i = 0; i < 32; ++i) {
            ans = (ans << 1) | (n & 1);
            n >>= 1;
        }
        return ans;
    }
};

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

############

class Solution:
  # @param n, an integer
  # @return an integer
  def reverseBits(self, n):
    ans = 0
    mask = 1
    for _ in range(32):
      ans <<= 1
      if mask & n:
        ans |= 1
      n >>= 1
    return ans

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
    }
}

190 - Reverse Bits

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