leetcode中等题

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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.

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class Solution {
public:
    vector<int> countBits(int num) {
        if(num<=0)
        {
            return vector<int>(1,0);
        }
        vector<int> c(num+1, 0);
        int i = 0;
        for(int i=1;i<=num;i++)
        {
            if(i%2==0)
            {
                c[i]=c[i>>1];
            }
            else
            {
                c[i]=c[i>>1]+1;
            }
        }
        return  c;
    }
};

JAVA

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