# 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
for _ in range(32):
ans <<= 1
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
}
}