[robinyoon-dev] WEEK 04 Solutions#2465
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robinyoon-dev merged 6 commits intoDaleStudy:mainfrom Mar 29, 2026
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junzero741
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Mar 28, 2026
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| var maxDepth = function (root) { | ||
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| //NOTE: 현재 노드의 최대 깊이는 왼쪽 자식 트리의 최대 깊이와 오른쪽 자식 트리의 최대 깊이 중 큰 값에 1(현재 노드)을 더한 값 | ||
| if (root === null) { | ||
| return 0; | ||
| } | ||
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| const leftDepth = maxDepth(root.left); | ||
| const rightDepth = maxDepth(root.right); | ||
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| return Math.max(leftDepth, rightDepth) + 1; |
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해설 써주신 것 좋습니다! 풀이도 깔끔하구요!
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| if(nums[currentIndex - 1] > nums[currentIndex]){ | ||
| result = nums[currentIndex]; | ||
| }else{ | ||
| continue; | ||
| } |
Contributor
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개인 차는 있겠지만, 저는 else 문이 중괄호랑 간격이 좀 있으면 가독성이 더 좋은 것 같더라구요!
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| let hasCharOnTheLeft = dfs(x - 1, y, nextIndex); | ||
| let hasCharOnTheRight = dfs(x + 1, y, nextIndex); | ||
| let hasCharAtTheTop = dfs(x, y + 1, nextIndex); | ||
| let hasCharAtTheBottom = dfs(x, y - 1, nextIndex); | ||
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| board[x][y] = tempChar; | ||
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| return hasCharOnTheLeft || hasCharOnTheRight || hasCharAtTheTop || hasCharAtTheBottom; |
Contributor
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has~ 로 4개의 변수를 만들기보단, 1개의 변수로 줄일 수 있을 것 같아요.
const found = dfs(r + 1, c, index + 1) ||
dfs(r - 1, c, index + 1) ||
dfs(r, c + 1, index + 1) ||
dfs(r, c - 1, index + 1);
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