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[Hoonjichoi] WEEK 02 solutions #2696
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,26 @@ | ||
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| /* | ||
| f(45) = f(44) + f(43) | ||
| f(44) = f(43) + f(42) | ||
| . | ||
| . | ||
| . | ||
| f(3) = f(2) + f(1) | ||
| f(2) = 2 | ||
| f(1) = 1 | ||
| */ | ||
| class Solution { | ||
| public int climbStairs(int n) { | ||
| if (n <= 2) return n; | ||
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| int prev = 1; | ||
| int result = 2; | ||
| int temp = 0; | ||
| for (int i = 2; i < n ; i++) { | ||
| temp = result; | ||
| result += prev; | ||
| prev = temp; | ||
| } | ||
| return result; | ||
| } | ||
| } |
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Contributor
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🏷️ 알고리즘 패턴 분석
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,39 @@ | ||
| import java.util.HashMap; | ||
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| public class hoonjichoi1 { | ||
| public boolean isAnagram(String s, String t) { | ||
| if (s.equals(t)) | ||
| return true; | ||
| if (s.length() != t.length()) | ||
| return false; | ||
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| HashMap<Character, Integer> map = new HashMap<>(); | ||
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| for (int i = 0; i < s.length(); i++) { | ||
| Character c = s.charAt(i); | ||
| if (map.containsKey(c)) { | ||
| int value = map.get(c) + 1; | ||
| map.put(c, value); | ||
| } else { | ||
| map.put(c, 1); | ||
| } | ||
| } | ||
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| for (int j = 0; j < t.length(); j++) { | ||
| Character c2 = t.charAt(j); | ||
| if (map.containsKey(c2)) { | ||
| int value = map.get(c2) - 1; | ||
| map.put(c2, value); | ||
| } else { | ||
| return false; | ||
| } | ||
| } | ||
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| for (Integer i : map.values()) { | ||
| if (i < 0) { | ||
| return false; | ||
| } | ||
| } | ||
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Comment on lines
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Contributor
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 이 반복문은 2번째 반복문에서 처리해줘도 좋을 듯 싶습니다..! |
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| return true; | ||
| } | ||
| } | ||
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🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
풀이 1:
Solution.climbStairs— Time: O(n) / Space: O(1)피드백: 상수 공간으로 이전 두 값만 유지하며 반복문으로 계산한다. 루프는 n-2회 실행되므로 선형 시간 복잡도다.
개선 제안: 현재 구현이 적절해 보입니다.
풀이 2:
hoonjichoi1.isAnagram— Time: O(n) / Space: O(k)피드백: 해시맵을 사용해 문자의 등장 횟수를 세고, 두 문자열의 차이를 확인한다. 최악의 경우 모든 문자를 한 번씩 처리한다.
개선 제안: 현재 구현이 적절해 보입니다.