[刷题]Counting Bits

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.

 

思路：

O(n*n)过于简单，不再赘述。为了实现O(n)，需要观察一下。一下描述用c(n)来代表n包含二进制1的个数。

• c(0) = 0, c(1) = 1;
• c(2) = 1, c(3) = c(1) + c(2) = c(1) + 1;
• c(4) = 1, c(5) = c(1) + c(4) = c(1) + 1, c(6) = c(2) + c(4) = c(2) + 1, c(7) = c(3) + c(4) = c(3) + 1;
• ……
• c(n) = c(n-m) + 1; m是小于等于n的最大2的整次幂

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> vec;
        int cur = 1;
        int next = 2;
        for(int i=0;i<=num;i++) {
            if (i < 2)
            {
                vec.push_back(i);
                continue;
            }
            if (i == next)
            {
                cur = next;
                next = next << 1;
            }
            if (i >= cur)
            {
                vec.push_back(vec[i-cur]+1);
            }
        }
        return vec;
    }
};