LeetCode[338] 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.

每遇到2的次方数就是把从vec[0]开始每个元素+1

class Solution {public:    vector<int> countBits(int num) {        vector<int> vec;        vec.push_back(0);        int k = 1;        for (int i = 1; i <= num; i++)        {            if (i == k)                k*= 2;            vec.push_back(vec[i - k/2] + 1);        }        return vec;    }};

一开始用的 pow(2,k)，但那样用时太长