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[std-freejia] week 02 solutions #1753

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Merged
merged 4 commits into from
Aug 3, 2025
Merged

[std-freejia] week 02 solutions #1753

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b4d25ab
feat: week02 climbing-stairs
std-freejia Jul 30, 2025
cfca352
feat: week02 3sum
std-freejia Jul 30, 2025
5b24489
feat: week02 valid-anagram
std-freejia Jul 30, 2025
c8d979a
feat: product of array except self
std-freejia Jul 31, 2025
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14 changes: 14 additions & 0 deletions valid-anagram/std-freejia.java
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@hj4645 hj4645 Aug 2, 2025 •
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이 문제는 맵을 하나만 선언해서 count를 빼주는 방식으로 풀었었는데, 2개를 선언해서 equals로 푸는 방법도 가독성 면에서 좋은 것 같습니다
2주차도 수고 많으셨습니다!😆
리뷰가 늦어 죄송합니다🥲

Map<Character, Integer> map = new HashMap<>();

        // s의 문자로 카운트 증가
        for (char c : s.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }

        // t의 문자로 카운트 감소
        for (char c : t.toCharArray()) {
            int count = map.getOrDefault(c, 0);
            if (count == 0) { // t에만 있는 문자가 있거나, 같은 문자가 더 많으면
                return false;
            }
            map.put(c, count - 1);
        }
        return true;

Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length()) return false;

Map<Character, Integer> mapS = new HashMap<>();
Map<Character, Integer> mapT = new HashMap<>();

for (int i = 0; i < s.length(); i++) {
mapS.put(s.charAt(i), mapS.getOrDefault(s.charAt(i), 0) + 1);
mapT.put(t.charAt(i), mapT.getOrDefault(t.charAt(i), 0) + 1);
}
return mapS.equals(mapT);
}
}