[Yn3-3xh] WEEK 05 Solutions #1402
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minji-go
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May 1, 2025
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14
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| if (min > prices[i]) { | ||
| min = prices[i]; | ||
| continue; | ||
| } | ||
| max = Math.max(max, prices[i] - min); |
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회고가 있는게 좋네요👍
저는 공간복잡도를 O(1)으로 만드는 부분이 어려웠던거 같아요!
최저가가 갱신되는 날은 판매하면 무조건 최대 이익이 아니다라는 관점에서 이해했는데,
이 부분이 이해하기 살짝 어렵게 느껴져서 Math.min()을 사용해서, 분기를 줄일 수도 있겠네요!
min = Math.min(min, prices[i]);
max = Math.max(max, prices[i] - min); continue가 없어서 max가 계산되어도, 값은 max 혹은 0이 될테니까요
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우와 생각하지 못했던 부분을 정성스럽게 피드백 해주셔서 감사합니다!
덕분에 분기를 줄여봤는데 더 깔끔해진 것 같아요 ㅎㅎ
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