# 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 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.

分析

背包问题，计算个数count， 当前dp\[v]由前面的dp\[v-k] +1推出来，这里难点是位运算

用c&(c-1)去掉最后一位1，就是原数缺了最后一个1

c - (c-1)\&c得到最后一个1，就是全部置空只有最后一个1的位置有1

```
class Solution:
    def countBits(self, num: int) -> List[int]:

        """
        :type num: int
        :rtype: List[int]
        """

        p = [0]*(num+1)

        for i in range(1,num+1):
            p[i] =p[i&(i-1)] + 1

        return p
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://nataliekung.gitbook.io/ladder_code/l4dong-tai-gui-hua/counting-bits.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.