经典算法-统计0~n之间的数字二进制的1的个数

经典算法-统计0~n之间的数字二进制的1的个数

1. 题目描述：
   给定一个数字n,统计0~n之间的数字二进制的1的个数，并用数字输出
2. 例子：
   当n = 5 时 返回的结果应该是：[0,1,1,2,1,2]
3. 要求：
   
   • 算法复杂度o (n)
   • 空间复杂度o (n)
4. 思路分析：
   
   • 根据题目的要求，时间和空间复杂度，明显是要用动态规划的方法
   • 得出推到公式：f(n) = 不大于f(n)的最大的2的次方+f(k),k一定是再前面出现的，用数组记录，直接查询
   • 举例 f(5) = f(4) + f(1), 注意2的次方都是一个1，而且是最高位，f(5) = 1 + f(1), f(6) = 1 + f(2)直到f(8) = 1

 public int[] countBits(int num){        int[] res = new int[num + 1];        int pow2 = 1;        int before = 1;        for(int i = 1; i <= num;i++){            if(i == pow2){                before = res[i]  = 1;                pow2 <<= 1;            }else{                res[i] = res[before] + 1;                before += 1;            }        }        return res;    }

原文描述：

Given a non negative integer number num. For every nubmers 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.