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： 观察可知： | 0 0 0 0 -- 0 | 0 0 0 1 -- 1 | 0 0 1 0 -- 1 | 0 0 1 1 -- 2 | 0 1 0 0 -- 1 | 0 1 0 1 -- 2 | 0 1 1 0 -- 2 | 0 1 1 1 -- 3 | 1 0 0 0 -- 1 | 1 0 0 1 -- 2 | 1 0 1 0 -- 2 | 1 0 1 1 -- 3 | 1 1 0 0 -- 2 | 1 1 0 1 -- 3 | 1 1 1 0 -- 3 | 1 1 1 1 -- 4 题解： #include <vector> std::vector<int> countBits(int num) { std::vector<int> result(num + 1); // extra 0 at beginning result[0] = 0; int z = 1; while((z << 1) <= num) { for(int i = 0; i < z; ++i) { result[i + z] = result[i] + 1; } z <<= 1; } for(int i = 0; z <= num; ++i, ++z) { result[z] = result[i] + 1; } return result; }