7주차 알고리즘 문제 풀이(1) - 송헌욱 - #32
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juneheel
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Aug 11, 2024
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| public static String solution(int N, String[] words) { | ||
| StringBuilder result = new StringBuilder(); | ||
| Set<String> uniqueWords = new HashSet<>(List.of(words)); |
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중복제거를 Set을 이용해하셨네요! 이문제에서는 왜 생각을 못했는지,, 그리고 Set에서 배열을 List.of()로 바로생성해주는것을 알게되었어요👍
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네! String 배열을 하나하나 확인하기 보다는 그냥 Set으로 바로 처리했어요! 새로운 걸 알게 되셨다니 다행입니다!
juneheel
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Aug 11, 2024
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| List<String> sortedWords = uniqueWords.stream() | ||
| .sorted((a, b) -> { | ||
| if (a.length() == b.length()) { // 문자열 길이가 같은 경우 | ||
| return a.compareTo(b); // 사전 순으로 | ||
| } else { // 문자열 길이가 다른 경우 | ||
| return Integer.compare(a.length(), b.length()); // 길이 비교 | ||
| } | ||
| }) | ||
| .collect(Collectors.toList()); |
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역시 stream과 람다식을 잘 활용해야겠다는 생각이 다시한번드네요.. 일일히 comparator내의 compare을 오버라이딩했는데, stream().sorted() 람다식으로 표현하니 가독성도 늘어나고 처리과정도 확 주는것같습니다👍
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stream을 활용하면 확실히 가독성, 코드 작성에 도움이 많이 되더라구요! 자바에서는 희망같습니다 ㅠ
juneheel
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Aug 11, 2024
| long[] result = new long[n]; | ||
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| for (int i = 0; i < n; i++) { | ||
| String rev = new StringBuilder(strNum[i]).reverse().toString(); |
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StringBuilder의 reverse()메서드 사용하신게 인상깊네요😊
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문제: BOJ5648. - 역원소 정렬
Arrays.sort(result);정렬은 nlogn의 시간복잡도를 가진다고 알고 있음.Arrays.sort()메서드 사용문제: BOJ1181 - 단어 정렬
정렬의 시간 복잡도는 n log n 으로 알고 있음.
문제: BOJ1822 - 차집합
정렬의 시간 복잡도는 n log n 으로 알고 있음.