[gmlwls96] Week7 #913
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[gmlwls96] Week7 #913
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obzva
approved these changes
Jan 23, 2025
| s.forEach { // s를 조회하면서 글자를 subStr에 담는다. | ||
| if (subStr.contains(it)) { // 단, 겹치는 글자가 있을경우 subStr의 len을 기록하고, 초기화 한다. | ||
| max = max(max, subStr.length) | ||
| subStr.delete(0, subStr.length) |
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이 로직이 오류를 유발합니다
답안 통과를 하지 못하고 있는데, 인지하고 계신가요?
| @@ -0,0 +1,43 @@ | |||
| class Solution { | |||
| // 시간 : O(N), 공간 O(N) | |||
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문제 설명에서 주어진 변수가 m, n (2차원 배열의 행, 열)이므로 복잡도 분석의 결과를 m, n을 이용해서 표현해주시거나 별도의 추가적인 설명을 곁들여주세요
| var answerRoot: ListNode? = null | ||
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| while (currentHead != null) { | ||
| answerRoot = ListNode(currentHead.`val`).apply { next = answerRoot } |
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answerRoot 뒤에 새로운 ListNode 인스턴스를 생성해서 추가하고 있으니까 공간 복잡도가 O(1)이 아닐 것 같아요
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| // dp 알고리즘. pathMap[y][x]로 올수 있는 경로의 수는 | ||
| // 위쪽(pathMap[y - 1][x]) + 왼쪽( pathMap[y][x - 1]) 경로의 수의 합이다. |
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pathMap[y][x] = pathMap[y - 1][x]) + 왼쪽( pathMap[y][x - 1]
이 점화식을 잘 보면 공간복잡도를 1차원으로 줄일 수 있다는 걸 알 수 있습니다
한 번 최적화해보시는 걸 추천드립니다~!
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@gmlwls96 longest-substring 문제는 틀린 답안을 수정하지도 않으셨는데 병합 진행하셨네요? |
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