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.
自己的实现方法，效率略低，开始时用memset初始化数组，结果怎么试都超时
class Solution { public: vector<int> countBits(int num) { vector<int> ret(num + 1, 0); for (int i = 1; i <= num; ++i) { int x = i; while (x) { ret[i] += x % 2; x /= 2; } } return ret; } }; 另外一种，就是不知道怎么推的 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; }
