comments | difficulty | edit_url | tags | ||
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Easy |
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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?
We can extract each bit of n
from the least significant bit to the most significant bit, and then place it in the corresponding position of ans
.
For example, for the (n & 1) << (31 - i)
to extract the n
and place it on the ans
, then right shift n
by one bit.
The time complexity is
class Solution:
def reverseBits(self, n: int) -> int:
ans = 0
for i in range(32):
ans |= (n & 1) << (31 - i)
n >>= 1
return ans
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 && n != 0; ++i) {
ans |= (n & 1) << (31 - i);
n >>>= 1;
}
return ans;
}
}
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t ans = 0;
for (int i = 0; i < 32 && n; ++i) {
ans |= (n & 1) << (31 - i);
n >>= 1;
}
return ans;
}
};
func reverseBits(n uint32) (ans uint32) {
for i := 0; i < 32; i++ {
ans |= (n & 1) << (31 - i)
n >>= 1
}
return
}
impl Solution {
pub fn reverse_bits(mut n: u32) -> u32 {
let mut ans = 0;
for i in 0..32 {
ans |= (n & 1) << (31 - i);
n >>= 1;
}
ans
}
}
/**
* @param {number} n - a positive integer
* @return {number} - a positive integer
*/
var reverseBits = function (n) {
let ans = 0;
for (let i = 0; i < 32 && n; ++i) {
ans |= (n & 1) << (31 - i);
n >>= 1;
}
return ans >>> 0;
};