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]`.

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.

Hint:

1. You should make use of what you have produced already.

2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.

3. Or does the odd/even status of the number help you in calculating the number of 1s?

## 题目大意：

1. 你应当利用已经生成的结果。

2. 将数字拆分为诸如 [2-3], [4-7], [8-15] 之类的范围。并且尝试根据已经生成的范围产生新的范围。

3.  数字的奇偶性可以帮助你计算1的个数吗？

## 解题思路：

`递推式：ans[n] = ans[n >> 1] + (n & 1)`
```//c++版本
class Solution
{
public：
vector<int>countBits(int num)
{
//一个数组有(0~num)即num+1个元素，初始化为0
vector<int> v1(num+1,0);
for(int i=1;i<=num;i++)
{
v1[i]=v1[i>>1]+(i&1);
}
}
}```

15 / 15 test cases passed.

Status: Accepted

Runtime: 124 ms

Submitted: 0 minutes ago

`递推式：ans[n] = ans[n - highbits(n)] + 1`

```highbits(n) = 1<<int(math.log(x,2))
math.log()不是c++/c的函数，java中有```

```highbits(7) = 4   （7的二进制形式为111）

highbits(10) = 8 （10的二进制形式为1010）```

`递推式：ans[n] = ans[n & (n - 1)] + 1`
```//c++版本
class Solution
{
public：
vector<int>countBits(int num)
{
//一个数组有(0~num)即num+1个元素，初始化为0
vector<int> v1(num+1,0);
for(int i=1;i<=num;i++)
{
v1[i]=v1[n&(n-1)]+1;
}
}
}```