338. 1的位数 Counting Bits

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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. 
   

规律：

i等于二的N次幂时，Hamming weight为 1.

HammingWeight(2) = 1

HammingWeight(3) = HammingWeight(2) + HammingWeight(1) = 2

......

HammingWeight(8) = 1

HammingWeight(9) = HammingWeight(8) + HammingWeight(1) = 2

HammingWeight(10) = HammingWeight(8) + HammingWeight(2) = 2

HammingWeight(11) = HammingWeight(8) + HammingWeight(2) = 3

static public int[] CountBits(int n)
        {
            int[] arr = new int[n + 1];
            arr[0] = 0;
            int pow2 = 2;
            int before = 1;
            for (int i = 1; i < n + 1; i++)
            {
                if (i == pow2)
                {
                    arr[i] = 1;
                    before = 1;
                    pow2 <<= 1;
                }
                else
                {
                    arr[i] = arr[before] + 1;
                    before++;
                }
            }
            return arr;
        }

338. 1的位数 Counting Bits

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原文地址：http://www.cnblogs.com/xiejunzhao/p/cce4e58a8ac2624fd987b8c7cf89355d.html