338. Counting Bits 数字的二进制中1的个数

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.

我得解答：
主要是观察到
1
10
11
100
101
110
111
1000
上述观察到1->10/11  10->100/101 11->110/111
即循环地在每个数后面加0、1可得接下来的数字。因此第i位就是第i/2位+（i%2）的值。以下为我的代码：

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

2.别人的代码，思路一样，就是算的过程不一样

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;
return ret;
}
};

3.别人的代码，实现不一样
An
easy recurrence for this problem is f[i] = f[i / 2] + i % 2.

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;
}