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 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.

Rust Solution

struct Solution;

impl Solution {
    fn count_bits(num: i32) -> Vec<i32> {
        let n = num as usize;
        let mut res = vec![];
        for i in 0..=n {
            res.push(i.count_ones() as i32);
        }
        res
    }
}

#[test]
fn test() {
    let num = 2;
    let res = vec![0, 1, 1];
    assert_eq!(Solution::count_bits(num), res);
}

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