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.

1、二进制方法

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ret(num+1, 0);
        for (int i = 1; i <= num; ++i)
            ret[i] = ret[i&(i-1)] + 1;  //i&(i-1)减掉了一个二进制1
        return ret;
    }
};

 

2、模板库方法

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> vec; 
        for(int i=0;i<=num;++i)
        {
            bitset<32> bs(i);
            vec.push_back(bs.count());
        }
        return vec;
    }
};