## 题目描述：

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)`

## Python代码：

``````class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
ans = [0]
for x in range(1, num + 1):
ans += ans[x >> 1] + (x & 1),
return ans
``````

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

`highbits(n) = 1<<int(math.log(x,2))`

```highbits(7) = 4   （7的二进制形式为111）

highbits(10) = 8 （10的二进制形式为1010）```

## Python代码：

``````class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
ans = [0]
for x in range(1, num + 1):
ans += ans[x - (1<<int(math.log(x,2)))] + 1,
return ans
``````

highbits运算可以不必每次都执行，用变量pow记录当前数字i的highbits

## Python代码：

``````class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
ans = [0]
pow = 1
for x in range(1, num + 1):
ans += ans[x - pow] + 1,
pow <<= x == pow
return ans
``````

`递推式：ans[n] = ans[n & (n - 1)] + 1`

## Python代码：

``````class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
ans = [0]
for x in range(1, num + 1):
ans += ans[x & (x - 1)] + 1,
return ans
``````

Pingbacks已关闭。

1. 大睿-SCTrojan 发布于 2016年3月21日 10:40 #

我的想法类似于Gray Code因为我发现[1]->[1], [2,3]->[1,2], [4,7]->[1,2,2,3], [8-15]->[1,2,2,3,2,3,3,4]，新的范围是上一个区间再加上上一个区间每个元素+1的结果，当然这个算法没有上面给出的好。

2. 在线疯狂 发布于 2016年3月21日 19:28 #

另一种解法GET√ :D

3. 张明锐 发布于 2016年7月19日 12:20 #

第二个解法的优化，应该是：pow <<= (x == (pow<<1))吧？