[선재] WEEK10 Solutions #537
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haklee
reviewed
Oct 19, 2024
| * brainstorming: | ||
| * preorder traverse | ||
| * | ||
| * n = length of root |
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혹시 length of root은 무엇을 의미하는 것일까요?
HC-kang
reviewed
Oct 19, 2024
| const dfs = (current) => { | ||
| const linkedNode = graph.get(current); | ||
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| if (memo[current] || !linkedNode || linkedNode.length === 0) return true; |
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이 부분은 옵셔널 체이닝을 사용하시면 조금 더 간결해질 것 같아요!
if (memo[current] || linkedNode?.length){ ... }
| * brainstorming: | ||
| * preorder traverse | ||
| * | ||
| * n = length of root |
There was a problem hiding this comment.
length of root는 전체 노드의 수를 말씀하시는걸까요?
| @@ -0,0 +1,43 @@ | |||
| /** | |||
| * @description | |||
| * queue의 특성을 활용하여 풀이 | |||
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이 부분에 대해서 혹시 추가 설명을 요청드려도 될까요?
| * space complexity: O(p) | ||
| */ | ||
| var canFinish = function (numCourses, prerequisites) { | ||
| const memo = Array.from({ length: numCourses + 1 }, () => false); |
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와 처음에는 메모가 과연 효과가 있을까..? 생각했는데 몇번을 돌려도 Beats 100%네요 👍
Sunjae95
commented
Oct 19, 2024
| const dfs = (current) => { | ||
| const linkedNode = graph.get(current); | ||
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| if (memo[current] || !linkedNode || linkedNode.length === 0) return true; |
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| if (answer === null) { | ||
| answer = lists[minIndex]; | ||
| tail = answer; | ||
| } else { | ||
| tail.next = lists[minIndex]; | ||
| tail = lists[minIndex]; | ||
| } |
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@HC-kang
매번 lists의 각 lists의 head중 가장 작은 index의 head를 바꿔주는방법이 queue의 선입선출을 연상되어서 적었습니다. 🙂
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