Description
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.
Solutions
1. Brute Force
# 1. Brute Force
# Time: O(num)
# Space: O(num)
class Solution:
def countBits(self, num: int) -> List[int]:
res = []
for i in range(num+1):
cnt = 0
while i != 0:
div, mod = divmod(i, 2)
if mod == 1:
cnt += 1
i = div
res.append(cnt)
return res
# 15/15 cases passed (300 ms)
# Your runtime beats 5.01 % of python3 submissions
# Your memory usage beats 5 % of python3 submissions (19.5 MB)
2. DP
状态转移矩阵：f[i] = f[i / 2] + i % 2.
# Time: O(num)
# Space: O(num)
class Solution:
def countBits(self, num: int) -> List[int]:
res = [0] * (num + 1)
for i in range(num + 1):
res[i] = res[i >> 1] + i % 2
return res
# 15/15 cases passed (88 ms)
# Your runtime beats 56.5 % of python3 submissions
# Your memory usage beats 5 % of python3 submissions (19.6 MB)