How to calculate the number of zeros in binary?

Hi I am making a method that can take an integer as a parameter and compute how many zeros its binary form has. So for example, if I have binaryZeros(44), its binary form is 101100. Therefore, binaryZeros(44) should return 3. However, I am making some errors and I cannot tell where it is coming from. I would appreciate it if someone can point out where I am making that error, or if my approach (logic) to this problem is good enough. Thank you!
My code is Below:

public static int binaryZeros(int n) {
int zeroCount = 0;
double m = n;
while (m >= 0.0) {
    m = m / 2.0;
    if (m == Math.floor(m)) {
        zeroCount++;
    } else {
        m = Math.floor(m);
    }
}
return zeroCount;
}

Below is a more concise way to solve this problem

public static int binaryZeros(int n) {
    int zeroCount = 0;
    
    // Run a while loop until n is greater than or equals to 1
    while(n >= 1)
    {
        /* Use modulo operator to get the reminder of division by 2 (reminder will be 1 or 0 as you are dividing by 2). 
           Keep in mind that binary representation is an array of these reminders until the number is equal to 1. 
           And once the number is equal to 1 the reminder is 1, so you can exit the loop there.*/ 
        if(n % 2 == 0)
        {
            zeroCount++;
        }

        n = n / 2;
    }
    return zeroCount;
}

Your approach is good, but I think there's a better way to do it. The Integer class has a static method that returns the binary of a number: Integer.toBinaryString(num) . This will return a String.
Then, you can just check if there are any 0 in that string with method that has a for loop and evaluating with an if:

public int getZeros(String binaryString){
int zeros = 0;
for(int i=0; i < binaryString.length; i++)
      if(binaryString.charAt[i].equals('0')
           zeros++;
return zeros;
}

I believe this would be a simpler option and it doesn't have any errors.

Once m == 0.0, it will never change, so your while loop will never stop.

Here is one way. It simply complements the integer reversing 1's and 0's and then counts the 1 bits. You should not be using floating point math when doing this.

• ~ complements the bits
• &1 masks the low order bit. Is either 1 or 0
• >>> shifts right 1 bit including sign bit.

System.out.println(binaryZeros(44) + " (" +Integer.toBinaryString(44) +")"); 
System.out.println(binaryZeros(-44) + " ("Integer.toBinaryString(-44)+")");
			
public static int binaryZeros(int v) {
	int count = 0;
	while (v != 0) {
        // count 1 bits 
        // of ~v
		count += (~v)&1;
		v >>>=1;
	}
	return count;
}

Prints

3 (101100)
4 (11111111111111111111111111010100)

</details>



# 答案5
**得分**: 0

```java
public static int getZeros(int num) {
    String str = Integer.toBinaryString(num);
    int count = 0;
    for (int i = 0; i < str.length(); i++) {
        if (str.charAt(i) == '0') count++;
    }
    return count;
}

toBinaryString() 方法返回整数参数的无符号二进制字符串表示。它接受一个整数数据类型的参数，并返回相应的二进制字符串。
然后，for循环在字符串中计算零的数量并返回它。

public static int getZeros(int num) {
		String str= Integer.toBinaryString(num);
		int count=0;
		for(int i=0; i<str.length(); i++) {
			if(str.charAt(i)=='0') count++;
		}
		return count;
	}

答案6

得分: 0

如果你从一个大于或等于 0 的数字 m 开始，无论你如何多次将它除以 2 或使用 Math.floor，它永远不会变为负数。循环应该在 m 达到 0 时停止，所以将条件改为 while (m > 0.0)。

请注意，你也可以使用内置的标准库方法来做同样的事情。例如，有一个方法可以返回数字中前导零的数量，还有一个方法可以返回数字中位为 1 的数量。通过同时使用这两个方法，你可以计算出非前导零的零的数量：

static int binaryZeros(int n) {
    return Integer.SIZE - Integer.numberOfLeadingZeros(n) - Integer.bitCount(n);
}

static int binaryZeros(int n) {
    return Integer.SIZE - Integer.numberOfLeadingZeros(n) - Integer.bitCount(n);
}

答案7

得分: 0

只要简单地使用Integer.bitCount(n)方法：

public static int binaryZeros(int n) {
    long val = n & 0xFFFFFFFFL;
    int totalBits = (int)(Math.log(val) / Math.log(2) + 1);
    int setBits = Long.bitCount(val);
    return totalBits - setBits;
}

public static int binaryZeros(int n) {
    long val = n & 0xFFFFFFFFL;
    int totalBits = (int)(Math.log(val) / Math.log(2) + 1);
    int setBits = Long.bitCount(val);
    return totalBits - setBits;
}