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).

Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

public class Solution {
public int[] countBits(int num) {
int[] ret = new int[num+1];
for (int i=0; i<=num; i++) {
ret[i] = count(i);
}
return ret;
}

private int count(int num) {
int cnt = 0;
while (num != 0) {
num &= (num - 1);//清除末尾1
cnt ++;
}
return cnt;
}
}