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].
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.
解题思路：
- 知道如何在Python中如何得到一个整数的二进制表示
- 在二进制表示中得到1的个数
#!usr/bin/env
# -*-coding:utf-8 -*-
class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
results = []
for i in range(num+1):
num = bin(i).count("1")
results.append(num)
return results
def countBits2(self, num):
"""
:type num: int
:rtype: List[int]
"""
res = [0]
for i in xrange(1, num + 1):
res.append(res[i >> 1] + (i & 1))
return res
if __name__ == "__main__":
sol = Solution()
numa = 5
print sol.countBits(numa)
numb = 9
print sol.countBits(numb)
print sol.countBits2(numb)
