191. Number of 1 Bits

191. Number of 1 Bits

一、題目

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?

 

二、程式作法

public class Solution
{
    public int HammingWeight(uint n)
    {
        uint counter = 0;
        while (n / 2 != 0)
        {
            counter += n % 2;
            n /= 2;
        }
        counter += n % 2;
        return (int)counter;
    }
}

/*不須要做 uint to int 轉換*/
/*
public class Solution
{
    public int HammingWeight(uint n)
    {
        int counter = 0;

        while (n / 2 != 0)
        {
            if (n % 2 == 1)
                counter++;
            n /= 2;
        }

        if (n % 2 == 1)
            counter++;

        return counter;
    }
}
*/

 

三、思路筆記