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.
对0到某个整数之间的所有数，求每个数二进制表示中1的个数。
思路是利用i/2和i只差一位的性质，dp[i] = dp[i >>1] + (i & 1)
代码：
class Solution { public: vector<int> countBits(int num) { vector<int> res(num+1, 0); for(int i = 1; i <= num; ++ i) res[i] = res[i & (i-1)] + 1; return res; } };