338. Counting Bits - M

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.

A:
 就是利用：每当一个bit 上赋值 1 之后。 是其前面的所有的重复。

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res(num+1,0);
        int end = 1; // end is exclusive
        while(end<=num)
        {
            for(int i = end; i< min(2*end, num+1); ++i)
                res[i] = 1+ res[i-end];
            end *= 2;
        }
        return res;
    }
};

Mistakes：