# 338.Counting Bits

**338.Counting Bits**

难度:Medium

> 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.
```

这题可以采用与[78.subsets](https://github.com/newdee/Leetcode/blob/master/78.subsets.md)相似的方法，循环，然后判断位数。

```
class Solution {
public:
    vector<int> countBits(int num) {
        int s= log(num)/log(2);
        if(pow(2,s) <num) s+=1;
        int N= 1;
        int count =1;
        vector<int> res(num+1, 0);
       for(int i=1;i<=num;i++)
       {
           int count =0;
           for(int j=0;j<=s;j++)
           if(i & (1<<j)) count++;
           res[i] = count;
       }
        return res;
    }
};
```

> 执行用时 :152 ms, 在所有 C++ 提交中击败了8.83%的用户\
> 内存消耗 :9.9 MB, 在所有 C++ 提交中击败了3.54%的用户

但是效率较低，可以采用动态规划的方法，对于一个数，通过该数的1/2的位数和该数末尾是否为1来确定该数的位数。

```
class Solution {
vector <int > counts={0,1,1,2,1};
public:
    vector<int> countBits(int num) {
      if(num <counts.size()) 
      {
          vector<int > res(counts.begin(), counts.begin()+num+1);
          return res;
      }
      for(int i=counts.size(); i<=num;i++)
      counts.push_back( counts[i/2] + i%2);
       return counts;
    }
};
```

> 执行用时 :84 ms, 在所有 C++ 提交中击败了67.28%的用户\
> 内存消耗 :10.8 MB, 在所有 C++ 提交中击败了5.19%的用户


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