README.md

Counting Bits

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.

Solution

很容易实现O(n * sizeof(num))的方法：

int countBit(int n)
{
	int sum = 0;
	while (n) {
		sum++;
		n &= (n - 1);
	}
	return sum;
}
int *countBits(int num, int *size)
{
	*size = num + 1;
	int* ans = (int *)malloc(sizeof(int) * *size);
	for (int i = 0; i <= num; ++i)
		ans[i] = countBit(i);
	return ans;
}

但题目要求O(n),我们可以使用迭代的方法，我们知道左移位除了最后一个1可能丢掉，不会改变其余位的1的个数，假设f(n)表示n的二进制1的个数，则：

• 若n为偶数，则f(n) = f(n >> 1).
• 若n为奇数，则f(n) = f (n >> 1) + 1. 其中后面的1是由于移位丢掉的。

int *countBits(int num, int *size)
{
	*size = num + 1;
	int* ans = (int *)malloc(sizeof(int) * *size);
	for (int i = 1; i <= num; ++i)
		ans[i] = ans[i >> 1] + (i & 1);
	return ans;
}