[LeetCode] 191. Number of 1 Bits

Problem

Write a function that takes an unsigned integer and returns the number of ‘1’ bits it has (also known as the Hamming weight). Note:

• Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It 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 3, the input represents the signed integer. -3.

Example 1:

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

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

Logic

1. 정수의 32비트 중 마지막 비트가 0인지 1인지 확인하고, 1이면 카운트를 세고, 그렇지 않으면 카운트를 세지 않는다.

2. 정수를 이용해 우측 시프트 비트 연산을 1만큼 수행한다.

3. 위 과정을 32번 거친 후 카운트를 반환한다.

   Code

   public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int cnt = 0;
        for (int i = 0; i < 32; i++) {
            cnt += n & 1;
            n >>= 1;
        } return cnt;
    }
   }
   

   Time Complexity

   • 정수 n의 비트 수를 n이라 가정할 때, 코드의 수행 시간을 좌우하는 것은 for반복문의 n번 계산 횟수이다. 따라서 본 솔루션은 O(n)의 시간 복잡도를 가진다.