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.
内建函数和挨个循环计算的方法都可行，不过题目明确不能用。。。
还是根据数来找规律
public int[] countBits(int num) {
int[] f = new int[num + 1];
for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
return f;
}
