[친환경사과] Week 6 #900
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[친환경사과] Week 6 #900
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mmyeon
reviewed
Jan 17, 2025
Comment on lines
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| if (element == ')' || element == ']' || element == '}') { | ||
| if (stack.isEmpty()) { | ||
| return false | ||
| } else if (stack.last() == '(' && element == ')') { | ||
| stack.removeAt(stack.lastIndex) | ||
| } else if (stack.last() == '[' && element == ']') { | ||
| stack.removeAt(stack.lastIndex) | ||
| } else if (stack.last() == '{' && element == '}') { | ||
| stack.removeAt(stack.lastIndex) | ||
| } else { | ||
| return false | ||
| } |
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괄호 짝인지 비교한 뒤 stack에서 제거 하는 로직이 반복되고 있어서,
괄호 짝을 객체나 Map에 정의한 뒤 키 값에 대응하는 값을 찾으면 중첩을 좀 더 줄일 수 있을 것 같아요
코틀린이 낯설어서 자바스크립트 코드 예시 첨부합니다 ~
const matchingBrackets: Record<string, string> = {
")": "(",
"}": "{",
"]": "[",
};There was a problem hiding this comment.
주석이 꼼꼼히 작성되어 있어서 리뷰하는 데 큰 도움이 되었어요.
더 나은 방법 찾으시는 모습이 멋지네요! 😃
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| fun maxArea02(height: IntArray): Int { | ||
| var maxValue = 0 | ||
| var left = 0 | ||
| var right = height.size - 1 | ||
| while (left <= right) { | ||
| val width = right - left | ||
| val containerHeight = Math.min(height[left], height[right]) | ||
| val area = width * containerHeight | ||
| maxValue = Math.max(maxValue, area) | ||
| if (height[left] < height[right]) { | ||
| left++ | ||
| } else { | ||
| right-- | ||
| } | ||
| } | ||
| return maxValue |
| * https://www.youtube.com/watch?v=TohdsR58i3Q 동영상 참조 | ||
| * trie 알고리즘을 사용한 문제 해결 | ||
| * 시간 복잡도: O(L * 26) | ||
| * -> addWord() 의 경우, 각 노드는 최대 26개의 노드를 가질 수 있으며 단어의 길이가 L이라면 O(L)의 시간 소요: O(L) |
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오! Trie 한 노드에서 가능한 자식 노드는 알파벳 개수 최대 26개로 제한되는군요!
중복된 알파벳은 기존 노드 재활용한다는 점 덕분에 새로 배워갑니다!
알려주셔서 감사합니다 👍
TonyKim9401
approved these changes
Jan 18, 2025
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