[강희찬] WEEK 8 Solution #502
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TonyKim9401
approved these changes
Oct 3, 2024
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| function dfs(node: _Node): _Node { | ||
| if (map.has(node.val)) return map.get(node.val)!; | ||
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| const newNode = new _Node(node.val); | ||
| map.set(node.val, newNode); | ||
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| for (let neighbor of node.neighbors) { | ||
| newNode.neighbors.push(dfs(neighbor)); | ||
| } | ||
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| return newNode; |
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결국 문제를 풀기 위해서는 재귀적으로 풀어야 하는데, 주어진 cloneGraph 자체를 재귀 함수로 사용하는 방법도 괜찮은것 같습니다!
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아래와 같은 형태로 해당 함수의 params를 변경한다면 cloneGraph를 직접 재귀함수로 사용해도 최적의 풀이가 가능할 것 같네요
function cloneGraph(node: _Node | null, map?: Map<number, _Node>): _Node | null {
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}| current = current.next; | ||
| } | ||
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| current.next = list1 || list2; |
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자스는 null이면 false를 반환하니 이게 가능하군요 :) 👍
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| maxCount = Math.max(maxCount, ++charCount[endCharIdx]); | ||
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| if (end - start + 1 - maxCount > k) { | ||
| charCount[s.charCodeAt(start) - A]--; | ||
| start++; |
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전위, 후위 연산을 사용 하시는 기준이 궁금해요 :)
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위 코드에서는 start와 end를 통해 윈도우의 크기를 줄이고 늘여가면서 maxCount를 계산하고있습니다
저는 일반적으로 연산 순서에 영향받지 않는다면 후위 연산을 선호하는데요, 별다른 이유는 없고 익숙해서 그렇습니다 ㅎㅎ
다만 전위 연산을 사용한 ++charCount[endCharIdx]의 경우에는 직전에 end가 +1로 증가한 상태이므로 maxCount와 비교하기 위해서는 Math.max()연산 이전에 증감을 시킬 필요가 있으므로 전위 연산자를 사용했습니다!
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