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.

思路:借鉴了java的解答.当num=5时

二进制   1个数
  0 --> 0
  1 --> 1
 10 --> 1
 11 --> 2
100 --> 1
101 --> 2

发现规律,后面的数可以利用前面的结果(动态规划,或者叫备忘录).例如数字5,二进制为101,可以将其分为两部分:
(1)最后一位,为1,可以用3%2获得
(2)剩余的数,为10(二进制),可以用5/2=2获得(即右移一位)
则5的bits等于2的bits加上1,即1+1=2.
推广可知,记i的Counting Bits为res[i],则
res[i] = res[i/2] + i%2
代码:

class Solution {
public:
    vector countBits(int num) {
        vector res(num+1); //初始化输出vector,长度为num+1,int型默认元素为0
        for (int i = 0; i <= num; i++) {  //一次遍历
            res[i] = res[i/2] + i%2;
        }
        return res;
    }
};