LeetCode题解：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.
思路：
观察可知：

| 0 0 0 0  --  0

| 0 0 0 1  --  1

| 0 0 1 0  --  1
| 0 0 1 1  --  2

| 0 1 0 0  --  1
| 0 1 0 1  --  2
| 0 1 1 0  --  2
| 0 1 1 1  --  3

| 1 0 0 0  --  1
| 1 0 0 1  --  2
| 1 0 1 0  --  2
| 1 0 1 1  --  3
| 1 1 0 0  --  2
| 1 1 0 1  --  3
| 1 1 1 0  --  3
| 1 1 1 1  --  4

题解：

#include <vector>

std::vector<int> countBits(int num) {
std::vector<int> result(num + 1);   // extra 0 at beginning
result[0] = 0;

int z = 1;
while((z << 1) <= num) {
for(int i = 0; i < z; ++i) {
result[i + z] = result[i] + 1;
}

z <<= 1;
}
for(int i = 0; z <= num; ++i, ++z) {
result[z] = result[i] + 1;
}

return result;
}