[재호] WEEK 06 Solutions #462
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obzva
reviewed
Sep 16, 2024
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| // 1. s의 길이만큼 순회를 하면서 | ||
| for (const char of s) { | ||
| // 2. 열린 괄호라면 stack에 짝지어진 닫힌 괄호를 저장 | ||
| // 3. 닫힌 괄호라면 stack에서 꺼낸 것과 동일한지 확인 | ||
| switch (char) { | ||
| case "(": | ||
| case "{": | ||
| case "[": | ||
| stack.push(map[char]); | ||
| break; | ||
| case "}": | ||
| case ")": | ||
| case "]": | ||
| if (stack.pop() !== char) { | ||
| return false; | ||
| } | ||
| } | ||
| } |
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괄호 문제에서 switch 쓸 생각은 못해봤는데 신선하네요!
sounmind
reviewed
Sep 18, 2024
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+76
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| const ROW = matrix.length; | ||
| const COLUMN = matrix[0].length; | ||
| // 우하좌상 순서 | ||
| const DIRECTION = [ |
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이렇게 상수로 선언한 덕분에 가독성이 확 올라가는 군요. 감사합니다
sounmind
approved these changes
Sep 18, 2024
bky373
approved these changes
Sep 19, 2024
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|
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| /** | ||
| * TC: O(N) | ||
| * SC: O(1) |
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addWord 는 단어 길이에 비례해 공간을 사용하는 것 같은데, 상수인 이유가 있을까용?
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제가 순간적으로 linked list로 착각했었네요 😅
O(N)으로 수정해두었습니다!
Comment on lines
+23
to
+40
| // 1. s의 길이만큼 순회를 하면서 | ||
| for (const char of s) { | ||
| // 2. 열린 괄호라면 stack에 짝지어진 닫힌 괄호를 저장 | ||
| // 3. 닫힌 괄호라면 stack에서 꺼낸 것과 동일한지 확인 | ||
| switch (char) { | ||
| case "(": | ||
| case "{": | ||
| case "[": | ||
| stack.push(map[char]); | ||
| break; | ||
| case "}": | ||
| case ")": | ||
| case "]": | ||
| if (stack.pop() !== char) { | ||
| return false; | ||
| } | ||
| } | ||
| } |
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괄호 문제에서 switch 쓸 생각은 못해봤는데 신선하네요!
obzva
reviewed
Sep 20, 2024
| } | ||
| } | ||
|
|
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| return Math.max(...longestLength); |
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마지막에 Math.max를 통해 longestLength배열의 최대값을 구하는 것보다 for 문 안에서 최대값을 갱신해주는 것이 연산량이 줄어들 것 같아요
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마지막 순회를 줄일수 있네요!
수정해서 반영했습니다!
감사합니다~!
obzva
approved these changes
Sep 20, 2024
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