Reverse Bits Solution
Leetcode Problem-190 [Type -> easy]:
Description:
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.
Video Explanation

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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.
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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.
Implementation in java:
Solutions
- C++
- Java
- Python
- JavaScript
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t result = 0;
for (int i = 0; i < 32; i++) {
result = (result << 1) | (n & 1);
n >>= 1;
}
return result;
}
};
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
return Integer.reverse(n);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNextInt()) {
int n = sc.nextInt();
Solution example = new Solution();
int result = example.reverseBits(n);
System.out.println(result);
}
sc.close();
}
}
class Solution:
def reverseBits(self, n: int) -> int:
res = 0
for _ in range(32):
res = (res << 1) | (n & 1)
n >>= 1
return res
var reverseBits = function(n) {
let result = 0;
for (let i = 0; i < 32; i++) {
result = (result << 1) | (n & 1);
n >>>= 1;
}
return result >>> 0;
};
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