# 题目

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].

1. 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
2. Space complexity should be O(n).
3. 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

1 1
2 10
3 11
4 100
8 1000
9 10001

list[i] = list[i%n]+1 //n = floor(log(i))

list[2] = list[2%2]+1

list[5] = list[5%4]+1

list[12] =list[12%8]+1

# 代码

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


# 更巧妙的代码

public int[] countBits(int num) {
int[] f = new int[num + 1];
for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
return f;
}