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 timeO(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.

结题思路：

二进制显示 十进制显示

000                0

001                1

010                2

011                3

100                4

101                5

110                6

111                7

第n个数的二进制中一个个数可以从前面已知的答案中寻找。当n%2==0的时候，n的二进制数为n/2的二进制数左移一位，则1的个数相同；当n%2!=0时，n的二进制数为n-1的二进制数加1。

class Solution {public:    vector<int> countBits(int num) {        vector<int> vc;        vc.push_back(0);        if (num >= 1){            vc.push_back(1);            for (int i = 2; i <= num; i++){                if (i % 2 == 0){                    vc.push_back(vc.at(i / 2));                }else{                    vc.push_back(vc.back() + 1);                }            }        }        return vc;    }};