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.

參考461題（http://blog.csdn.net/hiroshiten/article/details/72545807），利用n&(n-1)能夠去掉n中最右側一個1的性質，n包含1的個數等於n&(n-1)包含1的個數加1。

    vector<int> countBits(int num) {
        vector<int> ans(num+1,0);
        for (int i=1;i<num+1;i++)
        {
            ans[i]=ans[i&(i-1)]+1;
        }
        return ans;
    }