LeetCode[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).

一开始看此题，感觉无从下手。后来看了几遍，发现还可以，为了减少运算的次数，每次计算要用到上次的结果，这里的话也有些动态规划的思想。

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