[HoonDongKang] WEEK 03 solutions#1326
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HoonDongKang merged 5 commits intoDaleStudy:mainfrom Apr 19, 2025
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sora0319
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Apr 19, 2025
| return dpFunc(candidates, target); | ||
| } | ||
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| console.log(combinationSum([2, 3, 6, 7], 7)); |
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저는 백트레킹 방식을 썼는데 dp 방식으로도 할 수 있네요!!
백트레킹 보다 코드 이해는 쉽네요!
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| function optimizedFunc(s: string): number { | ||
| let prev2 = 1; | ||
| let prev1 = s[0] === "0" ? 0 : 1; | ||
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| for (let i = 1; i < s.length; i++) { | ||
| let curr = 0; | ||
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| const one = +s.slice(i, i + 1); | ||
| const two = +s.slice(i - 1, i + 1); | ||
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| if (one >= 1 && one <= 9) curr += prev1; | ||
| if (two >= 10 && two <= 26) curr += prev2; | ||
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| prev2 = prev1; | ||
| prev1 = curr; | ||
| } | ||
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| return prev1; | ||
| } | ||
| return optimizedFunc(s); |
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이 방식으로 공간 복잡도도 줄일 수 있네요
저는 이것도 dp를 사용해서 공간 복잡도가 O(n)인데 저장하지 않고 바로 비교하면 더 줄이게 되네요
| let count = 0; | ||
| while (n !== 0) { | ||
| count += n & 1; | ||
| n = n >>> 1; |
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논리 연산자를 사용하는 방법이 char를 하나씩 비교하는 것보다 더 직관적인 방법 같아요!
배우고 갑니다!
sora0319
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Apr 19, 2025
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