[LeetCode] 190. Reverse Bits - 문제풀이

Description

주어진 부호없는 32비트 정수를 뒤집어서 반환하는 문제입니다.

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 string00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is00111001011110000010100101000000.

Example 2:

Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation:The input binary string11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is10111111111111111111111111111111.

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?

Solution 1. bit Shift

public int reverseBits(int n) {
    int result = 0;
    for (int i = 0; i < 32; i++) {
        result += n & 1;
        n >>>= 1;   //  must do unsigned shift
        if (i < 31) //  for last digit, don't shift!
            result <<= 1;
    }
    return result;
}

result(0 000000)에 기존 값 n의 끝자리를 &1로 추출하여 제일 오른쪽에 더해주고 기존값은 부호없이 오른쪽으로 shift(>>>) 결과는 왼쪽으로 shift하여 역순으로 bit를 나열해줍니다.

Reference

Reverse Bits - LeetCode