# Question

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

 190	Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

Example 1:

Input:     00000010100101000001111010011100
Output:    00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596,
so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input:     11111111111111111111111111111101
Output:    10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293,
so return 3221225471 which its binary representation is 10111111111111111111111111111111.

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 signed integer type and should not affect your implementation, as the internal binary representation of the integer 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.

If this function is called many times, how would you optimize it?

Related problem: Reverse Integer

@tag-bit



# 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

Java

• 
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;
}
}
}

• // 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:
# @param n, an integer
# @return an integer
def reverseBits(self, n):
ans = 0