[leetcode 190] 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.

Follow up:

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

 

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

비트 뒤집는 문제 0->1을 바꾸는 게 아니라 비트 자체를 거꾸로 뒤집는다.

 

32비트가 최대 비트이므로 n의 0자리 비트부터 한개씩 바꿔준다. 

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 <<=1;
            result = result | (n & 1);
            n>>=1;
        }
        
        return result;
    }
}