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.
Hint:
- You should make use of what you have produced already.
- Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
规律：
i等于二的N次幂时，Hamming weight为 1.
HammingWeight(2) = 1
HammingWeight(3) = HammingWeight(2) + HammingWeight(1) = 2
......
HammingWeight(8) = 1
HammingWeight(9) = HammingWeight(8) + HammingWeight(1) = 2
HammingWeight(10) = HammingWeight(8) + HammingWeight(2) = 2
HammingWeight(11) = HammingWeight(8) + HammingWeight(2) = 3
static public int[] CountBits(int n){int[] arr = new int[n + 1];arr[0] = 0;int pow2 = 2;int before = 1;for (int i = 1; i < n + 1; i++){if (i == pow2){arr[i] = 1;before = 1;pow2 <<= 1;}else{arr[i] = arr[before] + 1;before++;}}return arr;}