Reverse Bits 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
Reverse Bits Solutions
✅Time: O(32)=O(1)
✅Space: O(1)
C++
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t ans = 0;
for (int i = 0; i < 32; ++i)
if (n >> i & 1)
ans |= 1 << 31 - i;
return ans;
}
};
Java
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
int ans = 0;
for (int i = 0; i < 32; ++i)
if ((n >> i & 1) == 1)
ans |= 1 << 31 - i;
return ans;
}
}
Python
class Solution:
def reverseBits(self, n: int) -> int:
ans = 0
for i in range(32):
if n >> i & 1:
ans |= 1 << 31 - i
return ans
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