[LeetCode] 0338. Counting Bits

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內容目錄

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:

It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

Python

				
					from typing import List


class Solution:
    def countBits(self, n: int) -> List[int]:
        result = []
        for i in range(n+1):
            result.append(str(bin(i)).split("0b")[1].count('1'))
        return result
    
print(Solution().countBits(5))

Easy

Python

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