描述
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.
解析
根据题意，可以找出规则如下：
| 数字 | 二进制中包含 1 的个数 |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 1 |
| 3 | 2 |
| 4 | 1 |
| 5 | 2 |
| 6 | 2 |
| 7 | 3 |
| 8 | 1 |
通过观察可以发现其中的规律，在 2**(n) 到 2**(n+1) 之间的数字的二进制中所包含的 1 的个数分别是 0 到 2**(n) 之间数字的二进制中所包含的 1 的个数加 1。时间复杂度为 O(N)，空间复杂度为 O(N)。
解答
class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
result = [0]*(num+1)
if num==0:
return result
result[0] = 0
result[1] = 1
i = 2
r = 2
while i < num+1:
if i < 2**r:
result[i] = result[i-2**(r-1)]+1
else:
r+=1
result[i] = result[i-2**(r-1)]+1
i+=1
return result
运行结果
Runtime: 68 ms, faster than 80.85% of Python online submissions for Counting Bits.
Memory Usage: 13.8 MB, less than 89.59% of Python online submissions for Counting Bits.
每日格言：逆境是达到真理的一条通路。
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