LeetCode 第 338 题 (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.

这道题也不难，计算一个数的 1 的位数时可以借用以前计算过的结果。下面是代码

class Solution {public:    vector<int> countBits(int num) {        vector<int> ret;        ret.push_back(0);        for(int i = 1; i <= num; i++)        {            int bits = ret[i >> 1] + (i & 1);            ret.push_back(bits);        }        return ret;    }};