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[아현] WEEK02 Solutions #349
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,27 @@ | ||
// time : O(n) | ||
// space : O(n) | ||
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class Solution { | ||
public boolean isAnagram(String s, String t) { | ||
char[] sToChar = s.toCharArray(); | ||
char[] tToChar = t.toCharArray(); | ||
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int[] charCount = new int[26]; | ||
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for(char c : sToChar) { | ||
charCount[c - 'a']++; | ||
} | ||
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for(char c : tToChar) { | ||
charCount[c - 'a']--; | ||
} | ||
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for(int i : charCount) { | ||
if(i != 0) { | ||
return false; | ||
} | ||
} | ||
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return true; | ||
} | ||
} | ||
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. 나머지 문제도 시간 되실 때 꼭 풀어보시면 좋을것 같습니다!! 배울게 많은 문제들이었던 것 같습니다=) |
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charArray가 반복문을 위해서만 사용된 것 같아서
charArray를 생성하지 않고 chatAt을 이용하면 공간복잡도를 O(1)으로 푸는 것도 가능할 것 같은데 어떻게 생각하실까요?
추가적으로 n 이 무엇을 의미하는 걸까요?
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n 은 문자열의 길이를 나타냅니다.
말씀하신 데로 charAt을 쓰면 sToChar와 tToChar 를 사용하지 않아도 되어서 공간복잡도가 O(1)로 끝날 수 있어서 좋은 선택지인 것 같습니다! 👍