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[inorrni] WEEK 02 solutions #1265
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,21 @@ | ||
class Solution { | ||
public boolean isAnagram(String s, String t) { | ||
// 속도: 443ms | ||
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// 길이가 같을 때 탐색 시작, 다르면 false | ||
if(s.length() == t.length()) { | ||
List<String> sList = new ArrayList<>(Arrays.asList(s.split(""))); | ||
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// t 각 문자에 대해 sList에서 제거 시도 | ||
for (String ch : t.split("")) { | ||
// remove(ch)는 해당 ch가 있으면 제거하고 true를 반환, 없으면 false를 반환 | ||
if (!sList.remove(ch)) { | ||
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. s, t를 모두 array로 만든 후, t를 for문돌려 문자 하나하나를 비교하며 제거하는 방식도 참신한것같아요! |
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return false; | ||
} | ||
} | ||
return true; | ||
} | ||
return false; | ||
} | ||
} | ||
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먼저 길이를 비교하는 것도 좋은 방법인것 같습니다!
좋은 아이디어 감사합니다 :)
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감사합니다~~~