题目描述：
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.
Hint:
- You should make use of what you have produced already.
- Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
- Or does the odd/even status of the number help you in calculating the number of 1s?
解题思路：
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
........
观察上面的情况，我们发现0,1,2-3,4-7,8-15为一组，且每组开头的1的位数都是1。每组其余的数都可以用本组开头的数加上另一个差值。且这两个数都已经在前面算过了。
代码如下：
public class Solution{
public int[] countBits(int num){
int[] res = new int[num + 1];
int pow = 1, k = 1;
res[0] = 0;
while(k <= num){
if(k == pow){
pow *= 2;
res[k++] = 1;
} else {
res[k] = res[pow / 2 ] + res[k - pow / 2];
k++;
}
}
return res;
}
}