Java [Leetcode 338]Counting Bits

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:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. 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;
	}
}