[강희찬] WEEK 6 Solution #468
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sounmind
reviewed
Sep 18, 2024
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| for (const char of s) { | ||
| if (pairs[char]) stack.push(char); | ||
| else if (char != pairs[stack.pop()!]) return false; |
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안녕하세요~ 희찬님!
!= 와 !==, == 와 ===를 혼용하시는 이유가 따로 있을까요?
저는 개인적으로 얕은 비교(==, !=)의 동작을 신경쓰지 않고 엄격한 비교만 일관적으로 하기 위해 깊은 비교를 사용하는 것을 선호하고 권장하는 편이에요.
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@sounmind 네 안녕하세요~ 저도 말씀하신 내용에 동의합니다.
최근 레거시 코드베이스에서 작업중이라 ==와 !=를 자주 사용하다 보니 습관적으로 얕은 비교를 사용하는 경우가 많아졌네요
덕분에 좀 더 의식하고 사용 할 수 있을 것 같습니다!
bky373
reviewed
Sep 19, 2024
| } | ||
| node = node[word[i]]; | ||
| } | ||
| return !!node['isEnd']; |
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오 js, ts 에는 이중부정(!!) 이란 게 있네요 여기에서 undefined 인 경우를 방지하기 위함이군요 ㅋㅋ 새롭네용 ㅎㅎㅎ
EgonD3V
approved these changes
Sep 20, 2024
| for (let i = 0; i < word.length; i++) { | ||
| if (word[i] === '.') { | ||
| for (const key in node) { | ||
| if (this.dfs(word.slice(i + 1), node[key])) { |
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string을 slice하는 로직은 그 길이만큼 시간 복잡도를 먹어서, 그 대신 index를 파라미터로 받아서 �시간복잡도를 개선할 수도 있습니다
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오 그렇게 할 수 있겠네요. 감사합니다~!
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