[Tony] WEEK 09 Solutions #518
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DaleSeo
approved these changes
Oct 12, 2024
| public List<List<Integer>> pacificAtlantic(int[][] heights) { | ||
| List<List<Integer>> output = new ArrayList<>(); | ||
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| if (heights.length == 0 || heights[0].length == 0) return output; |
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nit: 1 <= m, n <= 200이 제약 사항으로 명시되었으니 이 체크는 불필요할 것 같습니다.
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그렇네요! 습관적으로 적은것 같습니다. 주의해야겟네요 감사합니다!
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| if (nums[start] <= nums[mid]) { | ||
| min = Math.min(min, nums[start]); | ||
| start = mid + 1; | ||
| } else if (nums[mid] <= nums[end]) { | ||
| min = Math.min(min, nums[mid]); | ||
| end = mid - 1; | ||
| } |
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와, 이 알고리즘 굉장히 창의적인 것 같은데 모임 때 소개해주실 수 있으실까요?
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헉 ㅋㅋㅋ 다른 문제 이미 발표자료 준비 해 두긴 했는데, 간단하게라도 설명 할 수 있도록 준비해보겠습니다!
| @@ -0,0 +1,52 @@ | |||
| // TC: O(n) | |||
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시간 복잡도를 어떻게 분석하셨는지 설명 좀 부탁드리겠습니다.
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외부 while의 경우 s의 길이 만큼 반복하니 O(n)이 되구요.
내부 while의 경우 항상 실행되는 것이 아니라 각 문자가 슬라이딩 윈도우에 한번 추가되고, 최대 한번만 제거되기 때문에 시간 복잡도에 크게 영향을 미치지 않을거라 생각합니다.
정확하게 계산한다면 O(n + m) 이 될 것 같습니다!
wogha95
approved these changes
Oct 12, 2024
bky373
reviewed
Oct 12, 2024
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| while (right < s.length()) { | ||
| char c = s.charAt(right); | ||
| windowCounts.put(c, windowCounts.getOrDefault(c, 0) + 1); |
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(사소) Map.merge(..) 메서드를 사용하여 Map 접근 횟수도 줄이고 좀더 간단하게 작성할 수도 있습니다~
windowCounts.merge(c, 1, Integer::sum);
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