[Flynn] Week 08 #505
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[Flynn] Week 08 #505
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jdalma
reviewed
Oct 1, 2024
| while (lo < hi) { | ||
| int mid = lo + (hi - lo) / 2; | ||
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| if (can_make_valid_substring(s, mid, k)) lo = mid + 1; |
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바이너리 서치로 푸시다니 엄청 신선하네요
end 포인터를 한 칸씩 옮기면서 start 포인터를 움직이는 방식으로 풀었는데, 여기서는 "만들 수 있는 부분 문자열의 길이를 lo로 갱신" 하는 것이 핵심인가 보네요 ㅎㅎ 잘 봤습니다
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저도 아이디어가 안 떠올라서 리트코드의 에디토리얼을 참고했는데, 신선하더라구요 ㅎㅎ 감사합니다
이 풀이보다 시간복잡도 면에서 더 효율적인 풀이가 있긴 하지만요
jdalma
approved these changes
Oct 4, 2024
merge-two-sorted-lists/flynn.cpp
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| while (p != nullptr) { | ||
| node->next = p; | ||
| p = p->next; | ||
| node = node->next; | ||
| } | ||
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| while (q != nullptr) { | ||
| node->next = q; | ||
| q = q->next; | ||
| node = node->next; | ||
| } |
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제가 C++을 잘 모르지만.. 혹시 while문을 쓰지 않고 아래처럼 참조를 바로 담아주는 것은 어떨까요??
if (l1 != null) node.next = l1
if (l2 != null) node.next = l2There was a problem hiding this comment.
앗 그렇네요 ㅎㅎㅎ 좋은 지적 감사합니다 반영하겠습니다!
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| /** | ||
| * 풀이 2 | ||
| * - 풀이 1의 DP 전개 과정을 보면 우리한테는 DP 배열 두 행만 필요하다는 걸 알 수 있습니다 |
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오 unique-paths 문제처럼 최적화를 할 수 있었군요.. 잘 봤습니다!
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네 맞습니다 대부분의 2D 배열을 이용하는 DP 풀이는 이런 식의 공간 복잡도 최적화가 가능합니다 ㅎㅎ
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