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.

題目分析

對(duì)于一個(gè)給定的數(shù)字num, 給出[0,num]區(qū)間內(nèi)的每個(gè)數(shù)的二進(jìn)制表示所含的1的個(gè)數(shù)，用一個(gè)數(shù)組輸出結(jié)果。
嘗試給出符合以下條件的solution

1. T(n) = O(n)
2. S(n) = O(n)
3. 不使用C++或其他語(yǔ)言的內(nèi)置函數(shù)

解:
假定rst[num] 即為num的二進(jìn)制表示法中包含的1的個(gè)數(shù)，存在以下公式
rst[num] = rst[num^(num-1)] + 1
且rst[0] = 0;

Solution:

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;
}

T(n) = O(n)
S(n) = O(n)