Counting Bits

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.

思路

每次都在前面的基础上加上1
[0,1]
10,11，就是[0,1]加1
[100,101,110,111],就是前4个[0,1,10,11]加1
同理
所以只需使用前面的数据就行了

代码

class Solution {
public:
    vector<int> countBits(int num) {
        int i = 1;
        vector<int> result;
        if(num >= 0)
            result.push_back(0);
        if(num>=1)
            result.push_back(1);
        int end = pow(2,i+1);
        while(num>=end-1)
        {
            int begin = pow(2,i);
            int j=0;
            for(int k = begin;k<=end-1;k++)
            {
                result.push_back(result[j]+1);
                j++;
            }
            i++;
            end = pow(2,i+1);

        }

        int j=0;
        for(int k = pow(2,i);k<=num;k++)
        {
            result.push_back(result[j]+1);
            j++;
        }    

        return result;

    }
};

---

class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        min_num, max_num = 0, 1 
        arry = []
        arry.append(0) # 0 is the first element
        for i in range(1,num+1):
            if i < max_num:
                arry.append(arry[i-min_num]+1)
            else:
                arry.append(1)
                min_num = max_num
                max_num = max_num << 1
        return arry

从0,2,4,8分段 因为都是1
然后 9的末尾就是9-8 =1相同 8就是max_num