Leetcode--338. Counting Bits

题目

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

还可以从低位入手。1’的个数等于除了最低位之外的‘1’的个数加上最低位‘1’的个数，即ret[n] = ret[n>>1] + n%2，具体代码：

代码

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