[강희찬] WEEK 7 Solution #491
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DaleSeo
reviewed
Sep 25, 2024
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| while (right < s.length) { | ||
| if (MAP.has(s[right])) { | ||
| left = Math.max(MAP.get(s[right])! + 1, left); |
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와, Map을 사용해서 인덱스를 저장하니, left 인덱스를 하나씩 증가시키지 않고 한 번에 점프할 수가 있네요. 한 수 배워 갑니다! 🙇
bky373
reviewed
Sep 25, 2024
Comment on lines
+59
to
+69
| for (let i = 1; i < r; i++) { | ||
| if (matrix[i][0] === 0) { | ||
| matrix[i].fill(0); | ||
| } | ||
| } | ||
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| for (let j = 1; j < c; j++) { | ||
| if (matrix[0][j] === 0) { | ||
| matrix.forEach((row) => (row[j] = 0)); | ||
| } | ||
| } |
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(질문) 반복문을 하나로 합쳐 진행해도 좋을 것 같은데 두 개로 나누신 이유가 따로 있으실까요?
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넵 질문주셔서 감사합니다!
matrix의 가로, 세로 사이즈가 다를 수 있어서 분리 하였는데, 취향에 따라 아래와 같이 표현할 수는 있을 것 같습니다.
for (let i = 1; i < Math.max(r, c); i++) {
if (matrix[i]?.[0] === 0) matrix[i].fill(0);
if (matrix[0]?.[i] === 0) matrix.forEach((row) => (row[i] = 0));
}다만 지극히 제 주관적으로, 위 코드는 전체 라인 수는 줄어들지만 조건이나 처리가 다소 지저분하다고 느껴져서 두 개의 순회로 분리하였습니다!
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아 네 희찬님, 제가 생각했던 코드는 요런 방식이었습니다!
이 방법이 가독성이 크게 떨어지지는 않는 것 같아서 소소하게 코멘트 남기려 했습니다!
Suggested change
| for (let i = 1; i < r; i++) { | |
| if (matrix[i][0] === 0) { | |
| matrix[i].fill(0); | |
| } | |
| } | |
| for (let j = 1; j < c; j++) { | |
| if (matrix[0][j] === 0) { | |
| matrix.forEach((row) => (row[j] = 0)); | |
| } | |
| } | |
| for (let i = 1; i < r; i++) { | |
| for (let j = 1; j < c; j++) { | |
| if (matrix[i][0] === 0 || matrix[0][j] === 0) { | |
| matrix[i][j] = 0; | |
| } | |
| } | |
| } |
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