[LeetCode] 191. Number of 1 Bits

191. Number of 1 Bits

Easy

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?

 

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문제 풀이

문제 접근

1. 비트연산 문제인것 같다.

풀이

1. n 값을 오른쪽으로 쉬프트 하면서 체크한다.
2. 모든 비트 값을 확인한다.

• 더 빠른 버전은, n - 1 한 값을 and 연산을 해나간다. 
• 1 을 뺐을 경우, 현재 위치에서 1 인자리가 0 으로 변한다.
  0100 에서 1 을 빼면 0011 로 변하고, 이를 원래 값과 and 연산하면 0000 로 변한다. 
• 모든 비트 값을 비교해볼 필요가 없기 때문에 조금더 빠르지만 제출시에는 몇 ms 차이 안난다. 

소스 코드

class Solution:
    def hammingWeight(self, n: int) -> int:
        res = 0
        while n:
            if n & 1:
                res += 1
            n = n >> 1
        return res
        
# Faster Solution
class Solution:
    def hammingWeight(self, n: int) -> int:
        res = 0
        while n:
            res += 1
            n &= (n - 1)
        return res