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
• class Solution {public:    vector
            
              countBits(int num) {        vector
             
               count_v(num+1,0);        for(int i = 1; i <= num;i++)        {            int x = i;            int count=0;            while(x)            {                if((x & 1)==1) count++;                x >>= 1;            }            count_v[i] = count;        }        return count_v;    }};
             
            

  注意：1.计算二进制中1的个数，难点在于时间复杂度。有三种基本求二进制1个数的方法：

• （1）定理求解：取余数的方法，最慢。
• （2）基本法（雾）：值和1进行&操作，然后>>算术右移，仍然有两个循环。（即为本道题我的解法）
• （3）快速法：n和n-1进行&操作，可以消去二进制最右的1，仍然有两个循环。（在191题中的解法）
• 参考：http://www.cnblogs.com/graphics/archive/2010/06/21/1752421.html