Leetcode 338 Counting Bits

338. Counting Bits

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Question

Total Accepted: 7206  Total Submissions: 13074  Difficulty: Medium

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5 you should return [0,1,1,2,1,2].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

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Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

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  (E) Number of 1 Bits

这道题是191. Number of 1 Bits的升级版本，需要输出0-N内每个数的二进制1的个数，最终返回vector。

解法一：O(n * numBits(n))

这种解法的复杂度是 O(n * numBits(n)) 其中numBits(n)表示n的二进制表示1的位数。

【函数迭代的次数 = n中二进制1的个数】

代码如下：

int bitcount(unsigned int n)
{
	int count = 0;
	while (n)
	{
		count++;
		n &= (n - 1);
	}
	return count;
}

接下来就是遍历0-n内每个数，调用上面那个函数即可。

Solution如下(已AC 2016-03-25)：

int bitcount(unsigned int n)
{
	int count = 0;
	while (n)
	{
		count++;
		n &= (n - 1);
	}
	return count;
}

vector<int> countBits(int num) 
{
	vector<int> res;
	res.push_back(0);

	if (num <= 0) return res;

	else
	{
		for (int i = 1; i <= num; ++i)
		{
			res.push_back(bitcount(i));
		}
		return res;
	}
}