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 1:

Input: 2
Output: [0,1,1]
Example 2:

Input: 5
Output: [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.

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/counting-bits
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我的思路：

1. dp[i]表示數字 i 的 1 的個數

2. dp[i] = dp[i/2] + i%2;

AC

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

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