comments | difficulty | edit_url |
---|---|---|
true |
Easy |
Write a program to swap odd and even bits in an integer with as few instructions as possible (e.g., bit 0 and bit 1 are swapped, bit 2 and bit 3 are swapped, and so on).
Example1:
Input: num = 2(0b10) Output 1 (0b01)
Example2:
Input: num = 3 Output: 3
Note:
0 <= num <=
2^30 - 1- The result integer fits into 32-bit integer.
We can perform a bitwise AND operation between num
and 0x55555555
to get the even bits of num
, and then shift them one bit to the left. Then, we perform a bitwise AND operation between num
and 0xaaaaaaaa
to get the odd bits of num
, and then shift them one bit to the right. Finally, we perform a bitwise OR operation on the two results to get the answer.
The time complexity is
class Solution:
def exchangeBits(self, num: int) -> int:
return ((num & 0x55555555) << 1) | ((num & 0xAAAAAAAA) >> 1)
class Solution {
public int exchangeBits(int num) {
return ((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa)) >> 1;
}
}
class Solution {
public:
int exchangeBits(int num) {
return ((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa)) >> 1;
}
};
func exchangeBits(num int) int {
return ((num & 0x55555555) << 1) | (num&0xaaaaaaaa)>>1
}
function exchangeBits(num: number): number {
return ((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >>> 1);
}
impl Solution {
pub fn exchange_bits(num: i32) -> i32 {
let num = num as u32;
(((num & 0x55555555) << 1) | ((num & 0xaaaaaaaa) >> 1)) as i32
}
}
class Solution {
func exchangeBits(_ num: Int) -> Int {
let oddShifted = (num & 0x55555555) << 1
let evenShifted = (num & 0xaaaaaaaa) >> 1
return oddShifted | evenShifted
}
}