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;
    }
}