C--统计二进制中1的个数

Posted 2021-06-21 L_add

统计二进制中1的个数

题目内容：写一个函数返回参数二进制中1的个数
例如：
15
00001111
4个1

int count_one(size_t num)//size_t 无符号（方便接受负值）
{
	int count = 0;
	while (num)
	{
		if (num % 2 == 1)
			count++;
		num /= 2;
	}
	return count;
}
int main()
{
	int num;
	scanf("%d", &num);
	int ret = count_one(num);
	printf("%d\\n", ret);
	return 0;
}

优化：右移

int count_one(int n)
{
	int count = 0;
	for (int i = 0; i < 32; i++)
	{
		if (((n >> i) & 1) == 1)
			count++;
	}
}

int main()
{
	int num;
	scanf("%d", &num);
	int ret = count_one(num);
	printf("%d\\n", ret);
	return 0;
}

循环32次，再优化

int count_one(int n)
{
	int count = 0;
	while (n)
	{
		n = n & (n - 1);
		count++;
	}
	return count;
}
int main()
{
	int num;
	scanf("%d", &num);
	int ret = count_one(num);
	printf("%d\\n", ret);
	return 0;
}