LeetCode 338. Counting Bits

Description

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.

Analysis

找规律递推，效率较高的规律是，一个数字的中1的个数，必然是将其最低的那个1去掉后再加1。
看起来有点像废话，但通过这句话我们就找到了递推关系式。

Code

class Solution {public:    vector<int> countBits(int num) {        vector<int> result(num + 1);        result[0] = 0;        result[1] = 1;        for (int i = 2; i <= num; i++)            result[i] = result[i & (i-1)] + 1;        return result;    }};

Appendix

• Link: https://leetcode.com/problems/counting-bits/
• Run Time: 66ms