leetcode--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.

解题思路：二进制是一个非常特殊的运算，使用位运算，统计1的个数。

c++:

class Solution {
public:
    vector<int> countBits(int num) {
       std::vector<int> v;
        for(int i=0;i<=num;i++){
         bitset<32> num(i);//使用bitset<32> 可以进行位运算
         v.push_back(num.count());//将1的个数插入到向量容器中
        }
   return v;
    }
};

java版本：

public class Solution {
    public int[] countBits(int num) {
        int [] res=new int[num+1];
        for(int i=0;i<=num;i++){
            res[i]=res[i>>1]+(i&1);//移位进行计算
        }
        return res;
        }
}