LeetCode-338. 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.

这道题还是考察对整数的二进制的理解。我们列举会发现，当 i%2 == 0时，i 的二进制中 1 的个数变少，和 i/2 的整数相比，它的二进制只是多了一个0。看到这里，问题就迎刃而解了。以下是我的代码：

class Solution {public:    vector<int> countBits(int num) {        vector<int> num1(num+1);        num1[0] = 0;        for(int i = 1; i<= num;i++){            if(i%2 == 0){                num1[i] = num1[i/2];            }            else{                num1[i] = num1[i-1] + 1;            }        }        return num1;    }};