Easy
Given an integer n
, return an array ans
of length n + 1
such that for each i
(0 <= i <= n
), ans[i]
is the number of 1
’s in the binary representation of i
.
Example 1:
Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10
Example 2:
Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101
Constraints:
0 <= n <= 105
Follow up:
O(n log n)
. Can you do it in linear time O(n)
and possibly in a single pass?__builtin_popcount
in C++)?# @param {Integer} num
# @return {Integer[]}
def count_bits(num)
result = Array.new(num + 1, 0)
border_pos = 1
incr_pos = 1
(1..num).each do |i|
# when we reach pow of 2, reset border_pos and incr_pos
if incr_pos == border_pos
result[i] = 1
incr_pos = 1
border_pos = i
else
result[i] = 1 + result[incr_pos]
incr_pos += 1
end
end
result
end