# 338     Counting Bits

## 338. [Counting Bits](https://leetcode.com/problems/counting-bits/description/)

## 1. Question

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.

## 2. Implementation

**(1) DP**

思路: 通过观察，每个数的1bit个数等于这个数除以2的数的1bit个数，如果这个数是奇数的话，再这基础上再加上1

```java
class Solution {
    public int[] countBits(int num) {
        int[] dp = new int[num + 1];

        for (int i = 1; i <= num; i++) {
            dp[i] = dp[i / 2] + (i & 1);
        }
        return dp;
    }
}
```

## 3. Time & Space Complexity

**DP：**&#x65F6;间复杂度O(n), 空间复杂度O(n)