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[구문영(GUMUNYEONG)] WEEK 2 Solution #357
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db84238
valid-anagram
1ba0f04
Merge branch 'DaleStudy:main' into main
GUMUNYEONG 5bf4506
Merge remote-tracking branch 'upstream/main'
GUMUNYEONG 003671d
Merge remote-tracking branch 'upstream/main'
ac48e98
Merge branch 'main' of https://github.com/GUMUNYEONG/leetcode-study
265046d
valid-anagram 복잡도
4a91c11
Merge remote-tracking branch 'upstream/main'
GUMUNYEONG 68f8766
Merge branch 'main' of https://github.com/GUMUNYEONG/leetcode-study
GUMUNYEONG c690fd8
MOD : valid-anagram
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,32 @@ | ||
/** | ||
* @param {string} s | ||
* @param {string} t | ||
* @return {boolean} | ||
*/ | ||
var isAnagram = function (s, t) { | ||
const countHash = {}; | ||
let result = true; | ||
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if (s.length !== t.length) return false; | ||
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for (str_t of t) { | ||
countHash[str_t] ? countHash[str_t]++ : countHash[str_t] = 1; | ||
} | ||
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for (str_s of s) { | ||
if (countHash[str_s]) { | ||
countHash[str_s]--; | ||
} else { | ||
result = false; | ||
break; | ||
} | ||
} | ||
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return result; | ||
}; | ||
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// TC : O(n) | ||
// n(=s의 길이 = t의 길이) 만큼 반복 하므로 On(n) | ||
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// SC : O(n) | ||
// 최대크기 n(=s의 길이 = t의 길이)만큼인 객체를 생성하므로 공간 복잡도도 O(n) |
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요렇게 하시면 더 깔끔하고 효율적인 코드가 되지 않을까 생각이 들었습니다 :)
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앗 더 간단하게 표현할 수 있었네요..! 피드백 감사합니다! :>