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.
题目分析
对于一个给定的数字num, 给出[0,num]区间内的每个数的二进制表示所含的1的个数，用一个数组输出结果。
尝试给出符合以下条件的solution
- T(n) = O(n)
- S(n) = O(n)
- 不使用C++或其他语言的内置函数
解:
假定rst[num] 即为num的二进制表示法中包含的1的个数，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)
