Evan, Week 4 #85
Evan, Week 4 #85
Conversation
| /** | ||
| * Time Complexity: O(1) | ||
| * Reason: | ||
| * - n.toString(2): Converts the number to a binary string, which has a fixed length of up to 32 bits. This is O(1) because the length is constant. |
There was a problem hiding this comment.
Not sure if this built-in function would take a constant amount of time regardless of the input size. 🤔
There was a problem hiding this comment.
Thanks for feedback
Hmm... I thought this way because input size is fixed with 32 bits.
There was a problem hiding this comment.
The processing speed will vary depending on the number given, but no matter how large the number gets, 32 bits is the maximum, so it's a constant time in terms of time complexity. Of course, constant time complexity doesn't mean it's efficient.
If you don't think so, can you give me your example?
| var countBits = function (n) { | ||
| const result = new Array(n + 1).fill(0); | ||
|
|
||
| for (let i = 1; i <= n; i++) { | ||
| /** The number of 1's in i divided by 2, except last bit of i */ | ||
| const totalOnes = result[i >> 1]; | ||
| const lastBit = i & 1; | ||
|
|
||
| result[i] = totalOnes + lastBit; | ||
| } | ||
|
|
||
| return result; | ||
| }; |
There was a problem hiding this comment.
What an elegant solution! 🏅
| var hammingWeight = function (n) { | ||
| return n.toString(2).replace(/0/g, "").length; | ||
| }; |
There was a problem hiding this comment.
JS를 모르는 저로써는 꽤나 신기한 코드네요 ㅎㅎ
toString(2)가 2진수로 변환하는 거고, replace(/0/g, "")는 Regex로 0을 찾아서 빈 문자열로 변환하는 코드로 이해했는데 맞을까요?
There was a problem hiding this comment.
맞습니다!
이진수 연산을 도와주는 편한 메서드가 많이 있네요ㅎㅎ
No description provided.