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(nsizeof(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.

public class Solution {
    public int[] countBits(int num) {
        if(num == 0){
            return new int[]{0};
        }
        int dp[] = new int[num+1];
        dp[0] = 0;
        dp[1] = 1;
        for(int i = 2 ; i<=num; i++){
            if((i & 1) == 1){
                dp[i] = dp[i>>1] + 1;
            }else{
                dp[i] = dp[i>>1];
            }
        }
        return dp;
    }
}

你可能感兴趣的:(338. Counting Bits)